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;
template <typename T>
T abs(T x) {
  return x > 0 ? x : -x;
}
template <typename T>
T sqr(T x) {
  return x * x;
}
template <typename T>
ostream &operator<<(ostream &s, const vector<T> &x) {
  s << "[";
  for (auto it : x) {
    s << it << ", ";
  }
  s << "]";
  return s;
}
template <typename T>
ostream &operator<<(ostream &s, const set<T> &x) {
  s << "{";
  for (auto it : x) {
    s << it << ", ";
  }
  s << "}";
  return s;
}
template <typename U, typename V>
ostream &operator<<(ostream &s, const pair<U, V> &x) {
  s << "(" << x.first << ", " << x.second << ")";
  return s;
}
template <typename T>
bool chmax(T &x, const T &y) {
  if (x < y) {
    x = y;
    return true;
  }
  return false;
}
template <typename T>
bool chmin(T &x, const T &y) {
  if (x > y) {
    x = y;
    return true;
  }
  return false;
}
const int maxn = 1e6 + 3;
vector<int> g[maxn];
bool used[maxn];
void dfs(int v) {
  used[v] = true;
  for (auto to : g[v]) {
    if (!used[to]) {
      dfs(to);
    }
  }
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) {
    if (!g[i].empty()) {
      dfs(i);
      for (int j = 1; j <= n; j++) {
        if (!used[j] && !g[j].empty()) {
          cout << 0 << "\n";
          return 0;
        }
      }
    }
  }
  cout << m * 1ll * (m - 1) / 2;
  return 0;
}
