Discription

You are given an undirected graph, constisting of n vertices and m edges. Each edge of the graph has some non-negative integer written on it.

Let‘s call a triple (u,?v,?s) interesting, if 1?≤?u?<?v?≤?n and there is a path (possibly non-simple, i.e. it can visit the same vertices and edges multiple times) between vertices u

Calculate the sum over modulo 10⁹?+?7 of the values of s

The first line of the input contains two integers n and m (1?≤?n?≤?100?000, 0?≤?m?≤?200?000) — numbers of vertices and edges in the given graph.

The follow m lines contain three integers u_i, v_i and t_i (1?≤?u_i,?v_i?≤?n, 0?≤?t_i?≤?10¹⁸, u_i?≠?v_i) — vertices connected by the edge and integer written on it. It is guaranteed that graph doesn‘t contain self-loops and multiple edges.

Print the single integer, equal to the described sum over modulo 10⁹?+?7.

4 4
1 2 1
1 3 2
2 3 3
3 4 1

12

4 4
1 2 1
2 3 2
3 4 4
4 1 8

90

8 6
1 2 2
2 3 1
2 4 4
4 5 5
4 6 3
7 8 5

62

In the first example the are 6 interesting triples:

1. (1,?2,?1)
2. (1,?3,?2)
3. (1,?4,?3)
4. (2,?3,?3)
5. (2,?4,?2)
6. (3,?4,?1)

In the second example the are 12 interesting triples:

1. (1,?2,?1)
2. (2,?3,?2)
3. (1,?3,?3)
4. (3,?4,?4)
5. (2,?4,?6)
6. (1,?4,?7)
7. (1,?4,?8)
8. (2,?4,?9)
9. (3,?4,?11)
10. (1,?3,?12)
11. (2,?3,?13)
12. (1,?2,?14)

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=100005,ha=1e9+7;
int hd[maxn],to[maxn*4],ne[maxn*4],num,cnt[69][2],tot,n,m,ans;
ll val[maxn*4],ci[69],a[69],Xor[maxn];
bool v[maxn],can[69];

inline int add(int x,int y){ x+=y; return x>=ha?x-ha:x;}
inline void ADD(int &x,int y){ x+=y; if(x>=ha) x-=ha;}
inline void addline(int x,int y,ll z){
    to[++num]=y,ne[num]=hd[x],hd[x]=num,val[num]=z;
}

inline int C(int x){ return (x*(ll)(x-1)>>1)%ha;}

inline void update(ll x){
	for(int i=0;i<=60;i++) if(x&ci[i]) can[i]=1;
}

inline void ins(ll x){
	for(int i=60;i>=0;i--) if(x&ci[i]){
		if(!a[i]){ a[i]=x,tot++,update(x); return;}
		x^=a[i];
	}
}

void dfs(int x,int fa){
	v[x]=1;
	for(int i=hd[x];i;i=ne[i]) if(to[i]!=fa){
		if(!v[to[i]]) Xor[to[i]]=Xor[x]^val[i],dfs(to[i],x);
		else ins(Xor[x]^Xor[to[i]]^val[i]);
	}
	
	for(int i=0;i<=60;i++) cnt[i][(Xor[x]&ci[i])?1:0]++;
}

inline void solve(){
	for(int o=1;o<=n;o++) if(!v[o]){
        memset(cnt,0,sizeof(cnt));
        memset(a,0,sizeof(a)),tot=0;
        memset(can,0,sizeof(can));
		dfs(o,0);
        
        /*
		for(int i=0;i<=60;i++)
	        for(int j=1;j<=n;j++) cnt[i][(Xor[j]&ci[i])?1:0]++;
	    */
	    
    	for(int i=0,now;i<=60;i++){
            now=0;
            
            if(can[i]){
            	ADD(now,cnt[i][0]%ha*(ll)cnt[i][1]%ha);
            	ADD(now,add(C(cnt[i][0]),C(cnt[i][1]))%ha);
            	now=now*(ll)(ci[tot-1]%ha)%ha;
    		}
    		else ADD(now,ci[tot]%ha*(ll)cnt[i][0]%ha*(ll)cnt[i][1]%ha);
    		
    		ADD(ans,now*(ll)(ci[i]%ha)%ha);
    	}
    }
}

int main(){
	ci[0]=1;
	for(int i=1;i<=60;i++) ci[i]=ci[i-1]+ci[i-1];
	
	scanf("%d%d",&n,&m);
	int uu,vv; ll ww;
	for(int i=1;i<=m;i++){
	    scanf("%d%d%I64d",&uu,&vv,&ww);
		addline(uu,vv,ww),addline(vv,uu,ww);
    }
    
	solve();
	
	printf("%d\n",ans);
	return 0;
}

CodeForces - 724G Xor-matic Number of the Graph