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;
vector<int> adj[300111];
bool used[300111];
int par[300111];
int dep[300111];
int seg[300111];
long long pre[300111];
pair<int, int> find(int x, int y) {
  pair<int, int> ret = make_pair(y, y);
  while (x != y) {
    ret.first = min(ret.first, x);
    ret.second = max(ret.second, x);
    x = par[x];
  }
  return ret;
}
void DFS(int x) {
  used[x] = 1;
  for (int v : adj[x]) {
    if (v == par[x]) continue;
    if (!used[v]) {
      par[v] = x;
      dep[v] = dep[x] + 1;
      DFS(v);
    } else {
      if (dep[v] < dep[x]) {
        pair<int, int> cur = find(x, v);
        seg[cur.first] = cur.second - 1;
      }
    }
  }
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  for (int i = 1; i <= n + 2; i++) seg[i] = n;
  for (int i = 1; i <= n; i++) {
    if (!used[i]) {
      par[i] = i;
      DFS(i);
    }
  }
  for (int i = n; i >= 1; i--) seg[i] = min(seg[i], seg[i + 1]);
  for (int i = 1; i <= n; i++) {
    pre[i] = pre[i - 1] + (seg[i] - i + 1);
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, d;
    scanf("%d%d", &l, &d);
    int leva = l - 1, desna = d, mid, ans = l - 1;
    while (leva <= desna) {
      mid = (leva + desna) / 2;
      if (seg[mid] <= d) {
        ans = mid;
        leva = mid + 1;
      } else
        desna = mid - 1;
    }
    printf("%lld\n",
           pre[ans] - pre[l - 1] + 1LL * (d - ans) * (d - ans + 1) / 2LL);
  }
  return 0;
}
