You are given an undirected graph with n vertices. There are no edge-simple cycles with the even length in it. In other words, there are no cycles of even length that pass each edge at most once. Let's enumerate vertices from 1 to n. 

You have to answer q queries. Each query is described by a segment of vertices [l; r], and you have to count the number of its subsegments [x; y] (l ≤ x ≤ y ≤ r), such that if we delete all vertices except the segment of vertices [x; y] (including x and y) and edges between them, the resulting graph is bipartite.

Input

The first line contains two integers n and m (1 ≤ n ≤ 3·105, 1 ≤ m ≤ 3·105) — the number of vertices and the number of edges in the graph.

The next m lines describe edges in the graph. The i-th of these lines contains two integers ai and bi (1 ≤ ai, bi ≤ n; ai ≠ bi), denoting an edge between vertices ai and bi. It is guaranteed that this graph does not contain edge-simple cycles of even length.

The next line contains a single integer q (1 ≤ q ≤ 3·105) — the number of queries.

The next q lines contain queries. The i-th of these lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n) — the query parameters.

Output

Print q numbers, each in new line: the i-th of them should be the number of subsegments [x; y] (li ≤ x ≤ y ≤ ri), such that the graph that only includes vertices from segment [x; y] and edges between them is bipartite.

Examples

Input

6 6
1 2
2 3
3 1
4 5
5 6
6 4
3
1 3
4 6
1 6


Output

5
5
14


Input

8 9
1 2
2 3
3 1
4 5
5 6
6 7
7 8
8 4
7 2
3
1 8
1 4
3 8


Output

27
8
19

Note

The first example is shown on the picture below:

<image>

For the first query, all subsegments of [1; 3], except this segment itself, are suitable.

For the first query, all subsegments of [4; 6], except this segment itself, are suitable.

For the third query, all subsegments of [1; 6] are suitable, except [1; 3], [1; 4], [1; 5], [1; 6], [2; 6], [3; 6], [4; 6].

The second example is shown on the picture below:

<image>
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
vector<int> edge[300010], path;
int rmax[300010];
long long srmx[300010], isum[300010], result;
bool visited[300010], incycle[300010];
void dfs(int i, int p) {
  visited[i] = true;
  path.push_back(i);
  for (int j : edge[i]) {
    if (j == p) {
      continue;
    } else if (visited[j] && !incycle[j]) {
      int kmin = n, kmax = 1, plen = path.size();
      for (int k = plen - 1; k >= 0; k--) {
        kmin = min(path[k], kmin);
        kmax = max(path[k], kmax);
        incycle[path[k]] = true;
        if (path[k] == j) {
          break;
        }
      }
      rmax[kmin] = kmax - 1;
    } else if (!visited[j]) {
      dfs(j, i);
    }
  }
  path.pop_back();
}
int smid(int l, int r) {
  int start = l - 1, end = r, mid;
  while (start < end) {
    mid = (start + end + 1) >> 1;
    if (rmax[mid] <= r) {
      start = mid;
    } else {
      end = mid - 1;
    }
  }
  return start;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m;
  int a, b;
  srmx[0] = 0;
  isum[0] = 0;
  for (int i = 1; i <= n; i++) {
    rmax[i] = n;
    incycle[i] = false;
    visited[i] = false;
  }
  for (int i = 0; i < m; i++) {
    cin >> a >> b;
    edge[a].push_back(b);
    edge[b].push_back(a);
  }
  for (int i = 1; i <= n; i++) {
    if (!visited[i]) {
      dfs(i, 0);
    }
  }
  for (int i = n - 1; i >= 1; i--) {
    rmax[i] = min(rmax[i + 1], rmax[i]);
  }
  for (int i = 1; i <= n; i++) {
    srmx[i] = srmx[i - 1] + rmax[i];
    isum[i] = isum[i - 1] + i;
  }
  int q, l, r;
  cin >> q;
  while (q--) {
    cin >> l >> r;
    int m = smid(l, r);
    result = srmx[m] - srmx[l - 1] + (long long)(r - m) * r +
             (long long)(r - l + 1) - isum[r] + isum[l - 1];
    cout << result << endl;
  }
  return 0;
}
