-------------------------------------------------AC code-------------------------------------------------

#include<cstdio>

#ifdef WIN32
#   define outl "%I64d"
#else
#   define outl "%lld"
#endif

using namespace std;

inline int read(){
    int x = 0;
    bool fl = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9'){
        if(ch == '-')   fl = 0;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9'){
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return fl ? x : -x;
}

/*------------------------------------------------------------------------------------------------------------//
    若将双边看做两条边后，图存在欧拉路，则成立 
    分三种情况讨论： 
    1、自环都为双边：不考虑自环，剩下的点要构成欧拉路则必须有2个奇点，因为单边连的至少一个点是奇点，易知只有两条 
    单边连在同一个顶点上时才满足欧拉路，所以 ans += sum{C(d[i], 2)}
    2、有且仅有一个自环为单边：取任意一条剩下的非单边非自环作为单边即可满足构成欧拉路，所以 ans += loop*(m-loop)
    3、有两个自环为单边：图一定没有奇点，即可构成欧拉回路，所以 ans += C(loop, 2)，loop为自环数 
//------------------------------------------------------------------------------------------------------------*/

typedef long long LL;
const int MAXN = 100005;
int n, m, u, v, d[MAXN], loop, fa[MAXN];
LL ans;
bool b[MAXN];

int findfa(int x){
    if(fa[x] != x)  fa[x] = findfa(fa[x]);
    return fa[x];
}
inline void unionn(int x, int y){
    fa[findfa(y)] = findfa(x);
}

inline LL C(int x){//calulate C(x, 2)
    return 1ll*x*(x-1)>>1;
}

int main(){
    n = read(), m = read();
    for(int i = 1; i <= n; i++) fa[i] = i;
    for(int i = 1; i <= m; i++){
        u = read(), v = read();
        if(u == v)  loop++, b[u] = 1;
        else{
            d[u]++, d[v]++;
            if(findfa(u) != findfa(v))  unionn(u, v);
        }
    }
    int f;
    for(int i = 1; i <= n; i++)
        if(b[i] || d[i]){
            f = i;
            break;
        }
    f = findfa(f);
    for(int i = 1; i <= n; i++)
        if(findfa(i) != f && (b[i] || d[i]))
            return puts("0"), 0;
    for(int i = 1; i <= n; i++) ans += C(d[i]);//no loop is one_way edge
    ans += 1ll*loop*(m-loop);                  //one loop is one_way edge
    ans += 1ll*C(loop);                        //two loops are one_way edge
    printf(outl, ans);
    return 0;
}