12 12
5 1
5 11
1 11
8 11
8 9
5 12
6 9
7 11
9 3
9 10
4 12
10 2
78
1 1
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
1 11
1 12
2 2
2 3
2 4
2 5
2 6
2 7
2 8
2 9
2 10
2 11
2 12
3 3
3 4
3 5
3 6
3 7
3 8
3 9
3 10
3 11
3 12
4 4
4 5
4 6
4 7
4 8
4 9
4 10
4 11
4 12
5 5
5 6
5 7
5 8
5 9
5 10
5 11
5 12
6 6
6 7
6 8
6 9
6 10
6 11
6 12
7 7
7 8
7 9
7 10
7 11
7 12
8 8
8 9
8 10
8 11
8 12
9 9
9 10
9 11
9 12
10 10
10 11
10 12
11 11
11 12
12 12

1
3
6
10
15
21
28
36
45
55
65
76
1
3
6
10
15
21
28
36
45
55
66
1
3
6
10
15
21
28
36
45
55
1
3
6
10
15
21
28
36
45
1
3
6
10
15
21
28
36
1
3
6
10
15
21
28
1
3
6
10
15
21
1
3
6
10
15
1
3
6
10
1
3
6
1
3
1

#include <bits/stdc++.h>
using namespace std;
long long n, m, q;
long long a, b, tmp;
long long vis[300000 + 5];
vector<long long> path[300000 + 5];
stack<long long> cir;
long long p[300000 + 5], pre[300000 + 5];
void DFS(long long now, long long par) {
  vis[now] = tmp++;
  cir.push(now);
  for (long long i = 0; i < path[now].size(); i++) {
    if (!vis[path[now][i]])
      DFS(path[now][i], now);
    else if (path[now][i] != par && vis[path[now][i]] < vis[now]) {
      long long mmin = 300000 + 1, mmax = 0;
      while (cir.size()) {
        mmin = min(mmin, cir.top());
        mmax = max(mmax, cir.top());
        if (cir.top() == path[now][i]) {
          cir.pop();
          break;
        }
        cir.pop();
      }
      p[mmin] = mmax - 1;
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    cin >> a >> b;
    path[a].push_back(b);
    path[b].push_back(a);
  }
  for (long long i = 1; i <= n; i++) {
    if (!vis[i]) DFS(i, 0);
  }
  tmp = n;
  for (long long i = n; i >= 1; i--) {
    if (!p[i])
      p[i] = tmp;
    else
      tmp = min(tmp, p[i]);
  }
  for (long long i = 1; i <= n; i++) {
    pre[i] = max(pre[i - 1] + p[i] - i + 1, (long long)0);
  }
  cin >> q;
  for (long long i = 0; i < q; i++) {
    cin >> a >> b;
    tmp = upper_bound(p + a, p + 1 + n, b) - (p);
    if (tmp <= n) {
      cout << (b - tmp + 2) * (b - tmp + 1) / 2 + pre[tmp - 1] - pre[a - 1]
           << endl;
    } else {
      cout << pre[n] - pre[a - 1] << endl;
    }
  }
  return 0;
}
