100 10
87 73
15 71
14 33
1 4
20 80
65 67
36 36
25 4
22 21
7 97
 4 2
3 3
3 4
 5 4
3 3
4 4
5 5
1 2
 4 4
1 2
2 3
3 1
4 4
 4 3
1 1
2 3
3 4
 1000000 1
255765 255765
 4 3
1 2
2 3
4 4
 4 2
1 1
1 2
 3 3
1 1
2 2
3 3
 5 4
1 1
2 3
2 4
2 5
 4 5
1 4
3 4
1 2
2 3
2 4
 4 4
2 3
2 4
3 4
4 4
 10 9
8 10
3 10
2 8
9 3
4 8
1 10
7 9
5 4
7 3
 10 10
1 2
2 3
3 4
4 5
5 1
6 6
7 7
8 8
9 9
10 10
 5 5
1 4
3 4
1 2
2 3
2 4
 3 3
1 2
2 2
3 3
 5 3
1 1
2 2
3 3
 6 12
1 2
1 3
2 3
1 1
2 2
3 3
4 5
5 6
4 6
4 4
5 5
6 6
 4 4
1 2
2 3
1 3
4 4
 5 4
1 2
3 3
4 4
5 5
 7 13
6 7
2 7
3 7
4 3
5 2
1 4
7 7
3 1
5 5
3 3
1 5
2 6
6 6
 2 2
1 1
2 2
 5 5
1 1
2 2
3 3
4 4
5 5
 1 1
1 1
 7 3
1 1
3 3
6 6
 3 3
1 1
2 3
3 3

0 1 0 0 0 0 0 1 0 0 8 6 12 0 8 0 0 0 0 0 57 0 0 0 0 0
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 100;
int n, m, tot, father[N], t;
long long ans, cnt, d[N];
vector<int> e[N];
bool b[N];
void dfs(int now) {
  if (b[now]) return;
  b[now] = true;
  tot++;
  for (int i = 0; i < e[now].size(); i++) dfs(e[now][i]);
  return;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) father[i] = i;
  for (int i = 1, x, y; i <= m; i++) {
    scanf("%d%d", &x, &y);
    b[x] = b[y] = true;
    if (x == y)
      cnt++;
    else {
      d[x]++;
      d[y]++;
      e[x].push_back(y);
      e[y].push_back(x);
    }
  }
  for (int i = 1; i <= n; i++)
    if (b[i]) t++;
  dfs(1);
  if (tot != t) {
    printf("0");
    return 0;
  }
  for (int i = 1; i <= n; i++) ans += d[i] * (d[i] - 1) / 2;
  ans += cnt * (m - cnt) + cnt * (cnt - 1) / 2;
  printf("%lld\n", ans);
  return 0;
}
