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;
const int MX = 300200;
int n, m, q, i, j, x, y, a[MX], w[MX], z[MX], c[MX];
long long s[MX], r[MX];
vector<int> g[MX], l[MX];
void dfs(int i, int p, int d) {
  a[d] = i;
  w[i] = d;
  for (int j = 0; j < g[i].size(); j++) {
    int k = g[i][j];
    if (k == p || w[k] >= d) continue;
    if (w[k]) {
      int mx = i, mn = i;
      for (int e = w[k]; e <= d; e++) {
        mn = min(mn, a[e]);
        mx = max(mx, a[e]);
      }
      c[mx] = max(c[mx], mn);
    } else
      dfs(k, i, d + 1);
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (i = 0; i < m; i++) {
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  for (i = 1; i <= n; i++)
    if (w[i] == 0) dfs(i, 0, 1);
  s[1] = 1;
  for (i = 2; i <= n; i++) {
    c[i] = max(c[i], c[i - 1]);
    s[i] = s[i - 1] + i - c[i];
  }
  scanf("%d", &q);
  for (i = 1; i <= q; i++) {
    scanf("%d%d", &x, &z[i]);
    l[x].push_back(i);
  }
  for (i = 1; i <= n; i++)
    if (l[i].size()) {
      x = lower_bound(c + 1, c + n + 1, i) - c;
      for (j = 0; j < l[i].size(); j++) {
        y = l[i][j];
        if (x <= z[y])
          r[y] = s[z[y]] - s[x - 1] + (x - i) * (x - i + 1LL) / 2;
        else
          r[y] = (z[y] - i + 1LL) * (z[y] - i + 2LL) / 2;
      }
    }
  for (i = 1; i <= q; i++) printf("%lld\n", r[i]);
  return 0;
}
