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;
long long can[300005] = {0};
long long ans[300005] = {0};
vector<pair<long long, long long> > edge[300005];
bool haved[300005] = {0};
bool havee[300005] = {0};
long long all[300005];
long long what[300005];
void F(long long here, long long now) {
  long long i, a, b;
  if (haved[here]) {
    if (now == 0) return;
    a = here;
    b = here;
    for (i = all[here]; i < now; i++) {
      a = min(a, what[i]);
      b = max(b, what[i]);
    }
    can[b] = max(can[b], a);
  } else {
    haved[here] = 1;
    all[here] = now;
    what[now] = here;
    for (auto i : edge[here]) {
      if (!havee[i.second]) {
        havee[i.second] = 1;
        F(i.first, now + 1);
      }
    }
  }
}
int main() {
  long long N, M, i, a, b, K, l, r, Ans;
  scanf("%lld %lld", &N, &M);
  for (i = 0; i < M; i++) {
    scanf("%lld %lld", &a, &b);
    edge[a].push_back(make_pair(b, i));
    edge[b].push_back(make_pair(a, i));
  }
  for (i = 1; i <= N; i++) F(i, 0);
  for (i = 1; i <= N; i++) {
    can[i] = max(can[i], can[i - 1]);
    ans[i] = ans[i - 1];
    ans[i] += (i - can[i]);
  }
  scanf("%lld", &K);
  while (K--) {
    Ans = 0;
    scanf("%lld %lld", &a, &b);
    l = a;
    r = b;
    if (can[a] >= a - 1)
      printf("%lld\n", ans[b] - ans[a - 1]);
    else if (can[b] <= a - 1)
      printf("%lld\n", ((a - (a - 1)) + (b - (a - 1))) * (b - a + 1) / 2);
    else {
      while ((r - l) > 1) {
        if (can[(r + l) / 2] >= a - 1)
          r = (r + l) / 2;
        else
          l = (r + l) / 2;
      }
      printf("%lld\n", ans[b] - ans[r - 1] + (1 + r - a) * (r - a) / 2);
    }
  }
  return 0;
}
