Little boy Igor wants to become a traveller. At first, he decided to visit all the cities of his motherland — Uzhlyandia.

It is widely known that Uzhlyandia has n cities connected with m bidirectional roads. Also, there are no two roads in the country that connect the same pair of cities, but roads starting and ending in the same city can exist. Igor wants to plan his journey beforehand. Boy thinks a path is good if the path goes over m - 2 roads twice, and over the other 2 exactly once. The good path can start and finish in any city of Uzhlyandia.

Now he wants to know how many different good paths are in Uzhlyandia. Two paths are considered different if the sets of roads the paths goes over exactly once differ. Help Igor — calculate the number of good paths.

The first line contains two integers n, m (1 ≤ n, m ≤ 10⁶) — the number of cities and roads in Uzhlyandia, respectively.

Each of the next m lines contains two integers u and v (1 ≤ u, v ≤ n) that mean that there is road between cities u and v.

It is guaranteed that no road will be given in the input twice. That also means that for every city there is no more than one road that connects the city to itself.

Print out the only integer — the number of good paths in Uzhlyandia.

In first sample test case the good paths are:

• 2 → 1 → 3 → 1 → 4 → 1 → 5,
• 2 → 1 → 3 → 1 → 5 → 1 → 4,
• 2 → 1 → 4 → 1 → 5 → 1 → 3,
• 3 → 1 → 2 → 1 → 4 → 1 → 5,
• 3 → 1 → 2 → 1 → 5 → 1 → 4,
• 4 → 1 → 2 → 1 → 3 → 1 → 5.

There are good paths that are same with displayed above, because the sets of roads they pass over once are same:

• 2 → 1 → 4 → 1 → 3 → 1 → 5,
• 2 → 1 → 5 → 1 → 3 → 1 → 4,
• 2 → 1 → 5 → 1 → 4 → 1 → 3,
• 3 → 1 → 4 → 1 → 2 → 1 → 5,
• 3 → 1 → 5 → 1 → 2 → 1 → 4,
• 4 → 1 → 3 → 1 → 2 → 1 → 5,
• and all the paths in the other direction.

Thus, the answer is 6.

In the second test case, Igor simply can not walk by all the roads.

In the third case, Igor walks once over every road.

【分析】给你一个双向路径图（可能不连通且存在自环,同样的边不会出现两次），n个节点，m条边，要求划出一条路径，使得其中（m-2）条边每条边走两次，另外两条边每条边走一次，问有多少种不同的路径。

这题咋一看，像不像欧拉路，很像啊，欧拉路是划出一条路径经过所有的边一次且仅一次，而这个是选出(m-2)条边来走两次。那我们不妨将这(m-2)条边再建一条，比如，

u<-->v，那我们再建一条u<-->v。那么如果修改后的图能跑出一条欧拉路的话，那么就符合题意了，所以我们就枚举这些边。由于（m-2）可能比较大，那么我们就枚举这个2，即不增建的边。

我们判定欧拉路的依据是奇数度数节点只有两个或者没有。那么我们枚举每一条边，将此边不增，其他的每条边都扩增一条，那么此时除了这条边连接的两个端点（u!=v）

以外，其他的所有节点的度数都是偶数，这两个节点的度数都是奇数。等会，我们还差一条边，，，现在已经有两个节点度数是奇数了，那么我们要拆的下一条边就必须是与这两个节点相关的（可以自己想想为什么）那么这条边的贡献就是这两个节点所连得边的和-1了。还有就是如果你拆了某个自环的边，也是可以的。还有种情况就是你一开始选的是一个自环的边，那么你下一条要拆的随便那条边都行。

#include <cstdio>

#include <map>

#include <algorithm>

#include <vector>

#include <iostream>

#include <set>

#include <queue>

#include <string>

#include <cstdlib>

#include <cstring>

#include <cmath>

using namespace std;

typedef pair<int,int>pii;

typedef long long LL;

const int N=1e6+;

const int mod=1e9+;

int n,m,s,k,t,cnt;

bool vis[N];

vector<int>edg[N];

int loop=;

LL ans=;

int in[N];

struct man{

    int u,v;

}a[N];

void dfs(int u){

    vis[u]=;

    for(int i=;i<edg[u].size();i++){

        int v=edg[u][i];

        if(!vis[v])dfs(v);

    }

}

int main()

{

    int x,y,w,l,r;

    scanf("%d%d",&n,&m);

    for(int i=;i<=m;i++){

        scanf("%d%d",&x,&y);

        a[i].u=x;a[i].v=y;

        if(x!=y){

            edg[x].push_back(y);

            edg[y].push_back(x);

        }

        else loop++;

        in[x]++;in[y]++;

    }

    for(int i=;i<=n;i++)

        if(in[i]!=){

            dfs(i);

            break;

    }

    for(int i=;i<=n;i++){

        if(!vis[i]&&in[i]!=){

            puts("");

            return ;

        }

    }

    for(int i=;i<=m;i++){

        x=a[i].u;

        y=a[i].v;

        if(x!=y){

            ans+=(edg[x].size()-+edg[y].size()-+loop);

        }

        else {

            ans+=(m-);

        }

    }

    printf("%lld\n",ans/);

    return ;

}