題意：先給一棵具有n個節點的樹, 然後再給出m條邊, 問從樹上刪去一條邊, 再從m條邊中刪去一條邊, 把這個圖分成至少兩部分的方案數

做法：

我們知道再樹上每加一條邊樹上一定有環, 假設我們將在環上的樹邊的累加標記都加一, 表示被一條m中的邊所覆蓋, 然後我們在分析, 對於一顆樹我們刪去任意一條邊, 樹一定會被分成兩部分, 那麼如果被刪去的邊的累加標記>=2， 說明有>=2條m中的邊連線成環的樹邊中有它, 也就是樹分成的兩部分有至少2條邊還連線著, 這樣的話不管你怎麼選都不可能被劃分成兩部分, 方案數 + 0， 而被覆蓋了一次的了？ 那就只能刪去讓它成環的那條邊, 方案數+1, 而覆蓋0次的最脆弱, 只要斷開它, 一定會有兩部分, 方案數 + m, 所以做一次樹上的差分就好了， 統計好每條邊被覆蓋的次數， 然後對應的累加答案即可

///                 .-~~~~~~~~~-._       _.-~~~~~~~~~-.
///             __.'              ~.   .~              `.__
///           .'//                  \./                  \\`.
///        .'//                     |                     \\`.
///       .'// .-~"""""""~~~~-._     |     _,-~~~~"""""""~-. \\`.
///     .'//.-"                 `-.  |  .-'                 "-.\\`.
///   .'//______.============-..   \ | /   ..-============.______\\`.
/// .'______________________________\|/______________________________`.
//#pragma GCC optimize("Ofast")
#pragma comment(linker, "/STACK:102400000,102400000")
//#pragma GCC target(sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx) 
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)
#define S_1(x) scan_d(x)
#define S_2(x,y) scan_d(x),scan_d(y)
#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
#define mp make_pair
#define pb push_back
typedef long long LL;
typedef pair <int, int> ii;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e5+5;
//const int maxx=1e6+10;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;


int up[maxn][23];
//maxx[maxn][23];
int dep[maxn], dis[maxn];
int cnt, head[maxn];

int dp[maxn];
struct node {
        int to, next, w;
}e[maxn<<1];
void init() {
        me(head,-1); me(dis,0);
        me(up,0);  me(dep,0);
        cnt = 0;  //me(maxx, 0);
}
void add(int u, int v, int w) {
    e[cnt] = node{v, head[u], w};
    head[u] = cnt++;
}

void dfs(int u,int fa,int d)
{
    dep[u] = d + 1;
    for(int i = 1 ; i < 20 ; i ++) {
            up[u][i] = up[up[u][i-1]][i-1];
            //maxx[u][i] = max(maxx[up[u][i-1]][i-1], maxx[u][i-1]);
    }
    for(int i = head[u] ; ~i ; i = e[i].next) {
            int to = e[i].to;
            if(to == fa) continue;
            dis[to] = dis[u] + e[i].w;
            up[to][0] = u;
            //maxx[to][0] = e[i].w;
            dfs(to, u, d+1);
    }
}

int LCA_BZ(int u,int v) {
    int mx = 0;
    if(dep[u] < dep[v]) swap(u,v);
    int k = dep[u] - dep[v];
    for(int i = 19 ; i >= 0 ; i --) {
        if((1<<i) & k) {
        //  mx = max(mx, maxx[u][i]);
            u = up[u][i];
        }
    }
    if(u == v) return u;
    for(int i = 19 ; i >= 0 ; i --) {
        if(up[u][i] != up[v][i]){
        //  mx = max(mx, maxx[u][i]);
        //  mx = max(mx, maxx[v][i]);
            u = up[u][i];
            v = up[v][i];
        }
    }
//     return max(mx, max(maxx[u][0], maxx[v][0]));
    return up[u][0];
}
void dfs1(int u,int fa) {
	for(int i=head[u];~i;i=e[i].next) {
		int v=e[i].to;
		if(v==fa) continue;
		dfs1(v,u);
		dp[u]+=dp[v];
	}
}
int n,m;
void solve() {
	W(s_2(n,m)!=EOF) {
		init();
		me(dp,0);
		FOr(1,n,i) {
			int x,y;
			s_2(x,y);
			add(x,y,1);
			add(y,x,1);
		}
		dfs(1,-1,0);
		FOR(1,m,i) {
			int x,y;
			s_2(x,y);
 			dp[x]++;
 			dp[y]++;
 			dp[LCA_BZ(x,y)]-=2;
 		}
 		dfs1(1,-1);
 		LL ans=0;
 		FOR(2,n,i) {
 			if(dp[i]==0) {
 				ans+=m;
 			}
 			if(dp[i]==1) {
 				ans+=1;
 			}
 		}
 		print(ans);
 	}
 }
 		
int main() {
    //freopen( "1.in" , "r" , stdin );
    //freopen( "1.out" , "w" , stdout );
    int t=1;
    //init();
    //s_1(t);
    for(int cas=1;cas<=t;cas++) {
        //printf("Case #%d: ",cas);
        solve();
    }
}