CodeForces788B 欧拉路

B. Weird journey

time limit per test

 2 seconds

memory limit per test

 256 megabytes

input

 standard input

output

 standard output

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.

Input

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.

Output

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

Examples

input

5 4
1 2
1 3
1 4
1 5

output

6

input

5 3
1 2
2 3
4 5

output

0

input

2 2
1 1
1 2

output

1

Note

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.

题意：

　　有n个点，m条边，问如果选择m-2条边走2次，2条边只走1次，一共有多少种走法(当且仅当只走一次的边的集合不相同时，走法才不相同)

题解：

　　可以这样想，其它边走两次，就是说，一次去，一次回来。

　　如果图不连通，那么绝对是0，每条边经过两次，遍历整个图，那是局对可行的，

　　因为这样就是边重复走一次，求欧拉回路，那是绝对可以的，去在回来不久行了，只要连通随便走。

　　这样两条边只经过一次，如果有自环，自环多余两个随便选两个都可以，

　　选一个自环其它随便选一条边也可以，

　　不然，就只能两条边同一端点才行，统计答案。

#include<cstring>
#include<algorithm>
#include<cmath>
#include<iostream>
#include<cstdio>
#define ll long long
#define N 1000007
using namespace std;

int n,m,t;
ll du[N];
int zh[N];
bool vis[N];
int cnt,head[N],next[N*2],rea[N*2];

void add(int u,int v)
{
    next[++cnt]=head[u];
    head[u]=cnt;
    rea[cnt]=v;
}
void dfs(int u)
{
    vis[u]=true;
    for (int i=head[u];i!=-1;i=next[i])
    {
        int v=rea[i];
        if (!vis[v]) dfs(v);
    }
}
int main()
{
    memset(head,-1,sizeof(head));
    scanf("%d%d",&n,&m);
    for (int i=1,x,y;i<=m;i++)
    {
        scanf("%d%d",&x,&y);
        add(x,y),add(y,x);
        if (x==y)
        {
            t++;
            zh[x]++;
            continue; 
        }
        du[x]++,du[y]++;
    }
    for (int i=1;i<=n;i++)
        if (du[i])
        {
            dfs(i);
            break;
        }
    for (int i=1;i<=n;i++)
        if (!vis[i])
        {
            if (du[i]||zh[i])
            {
                printf("0
");
                return 0;
            }
        }
    ll ans=0;
    ans+=(ll)t*(t-1)/2;//自环随便选。
    ans+=(ll)t*(m-t);//其它点随便选、
    for (int i=1;i<=n;i++)
        if (du[i]>=2) ans+=(ll)du[i]*(du[i]-1)/2;//表示一条边进，一条边出。
    printf("%lld",ans); 
}