E. Minimal Diameter Forest

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a forest — an undirected graph with nn vertices such that each its connected component is a tree.

The diameter (aka "longest shortest path") of a connected undirected graph is the maximum number of edges in the shortest path between any pair of its vertices.

You task is to add some edges (possibly zero) to the graph so that it becomes a tree and the diameter of the tree is minimal possible.

If there are multiple correct answers, print any of them.

Input

The first line contains two integers nn and mm (1≤n≤10001≤n≤1000, 0≤m≤n−10≤m≤n−1) — the number of vertices of the graph and the number of edges, respectively.

Each of the next mm lines contains two integers vv and uu (1≤v,u≤n1≤v,u≤n, v≠uv≠u) — the descriptions of the edges.

It is guaranteed that the given graph is a forest.

Output

In the first line print the diameter of the resulting tree.

Each of the next (n−1)−m(n−1)−m lines should contain two integers vv and uu (1≤v,u≤n1≤v,u≤n, v≠uv≠u) — the descriptions of the added edges.

The resulting graph should be a tree and its diameter should be minimal possible.

For m=n−1m=n−1 no edges are added, thus the output consists of a single integer — diameter of the given tree.

If there are multiple correct answers, print any of them.

Examples

input

Copy

4 2
1 2
2 3

output

Copy

2
4 2

input

Copy

2 0

output

Copy

1
1 2

input

Copy

3 2
1 3
2 3

output

Copy

2

Note

In the first example adding edges (1, 4) or (3, 4) will lead to a total diameter of 3. Adding edge (2, 4), however, will make it 2.

Edge (1, 2) is the only option you have for the second example. The diameter is 1.

You can't add any edges in the third example. The diameter is already 2.

題意：給你一個含n(n<=1000)個節點、m(<n)條邊的無向無環無重邊圖（森林），讓你新增邊將其連線成一棵樹，使樹的直徑最小，輸出最後樹的最短直徑，和你新增的邊。

其實思路很好想，明顯是求每顆樹的直徑的鏈的中點，然後用最長的那棵樹的直徑的中點連線其他樹的中點。（想想是不是）

關鍵是怎麼寫。。。

理清思路，先求出每個連通塊的點、中心，同時求出最長直徑的樹及其連通塊編號，然後新增邊構成大樹，然後再求一遍直徑。這個過程可以用若干個dfs來實現，注意細節。

程式碼：

#include<bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=1010;
vector<int>vc[maxn],cv;
set<int>st;
set<int>::iterator it,itt;
int n,m,k,id;
int ans,tmp,cnt,ct,lid;
int ansid,ansl;
int vis[maxn],idx[maxn],du[maxn];
int in[maxn];
void dfs(int u,int fa)
{
    vis[u]=1;
    st.insert(u);
    in[u]=1;
    for(int i=0;i<vc[u].size();i++)
    {
        int v=vc[u][i];
        if(v!=fa) dfs(v,u);
    }
}
void findc()
{
    vector<int>vv;
    for(it=st.begin();it!=st.end();)
    {
        itt=it;
        it++;
        int u=*itt;
        if(du[u]==1)
        {
            vv.push_back(u);
            in[u]=0;
            if(st.size()>1)st.erase(u);
        }
    }
    //cout<<(*st.begin())<<endl;
    while(st.size()>2)
    {
        //cout<<" I "<<(*it)<<endl;
        vector<int>v1;
        for(int i=0;i<vv.size();i++)
        {
            int u=vv[i];
            for(int j=0;j<vc[u].size();j++)
            if(in[vc[u][j]]){
                if(--du[vc[u][j]]==1)
                {
                    v1.push_back(vc[u][j]);
                    st.erase(vc[u][j]);
                    in[vc[u][j]]=0;
                }
            }
        }
        vv=v1;
    }
    //cout<<" * "<<(*st.begin())<<endl;
    cv.push_back(*st.begin());
}
void dfs2(int u,int fa,int dep)
{
    if(dep>ans)
    {
        ans=dep;
        lid=u;
    }
    for(int i=0;i<vc[u].size();i++)
    {
        int v=vc[u][i];
        if(v==fa) continue;
        dfs2(v,u,dep+1);
    }
}
int findzj(int u=1,int idd=1)
{
    ans=-1;
    dfs2(u,0,0);
    ans=-1;
    dfs2(lid,0,0);
    if(idd&&ans>ansl)
    {
        ansl=ans;
        ansid=id;
    }
    return ans;
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=0;i<m;i++)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        vc[u].push_back(v);
        vc[v].push_back(u);
        du[u]++;
        du[v]++;
    }
    id=1;
    ansl=-1;
    for(int i=1;i<=n;i++)
    if(!vis[i]) {
        st.clear();
        for(int j=1;j<=n;j++)
        in[j]=0;
        dfs(i,0);
        findc();
        findzj(i);
        id++;
    }
    //cout<<ansid<<" "<<ansl<<" "<<cv.size()<<endl;
    for(int i=0;i<cv.size();i++)
    if(i==ansid-1) continue;
    else {
        int u=cv[i];
        int v=cv[ansid-1];
        vc[u].push_back(v);
        vc[v].push_back(u);
    }
    printf("%d\n",findzj(1,0));
    for(int i=0;i<cv.size();i++)
    if(i==ansid-1) continue;
    else {
        int u=cv[i];
        int v=cv[ansid-1];
        printf("%d %d\n",u,v);
    }
    return 0;
}