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 MAXN = 300010;
int n, m, q, col[MAXN], dad[MAXN], from[MAXN];
long long psum[MAXN], sol[MAXN];
vector<int> e[MAXN];
vector<int> idx[MAXN], lb[MAXN];
int bio[MAXN];
void dfs(int x) {
  bio[x] = 1;
  for (int y : e[x])
    if (bio[y] && bio[y] != 2) {
      if (dad[x] == y) continue;
      int t = x;
      int maxi = -MAXN;
      int mini = MAXN;
      while (true) {
        maxi = max(maxi, t);
        mini = min(mini, t);
        if (t == y) break;
        t = dad[t];
      }
      from[maxi] = mini + 1;
    } else if (!bio[y]) {
      dad[y] = x;
      dfs(y);
    }
  bio[x] = 2;
}
long long sum(int i, int j) {
  return (i == 0) ? psum[j] : psum[j] - psum[i - 1];
}
int main(void) {
  scanf("%d %d", &n, &m);
  for (int i = (0); i < (m); ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    --u;
    --v;
    e[u].push_back(v);
    e[v].push_back(u);
  }
  scanf("%d", &q);
  for (int i = (0); i < (q); ++i) {
    int a, b;
    scanf("%d %d", &a, &b);
    --a;
    --b;
    lb[b].push_back(a);
    idx[b].push_back(i);
  }
  for (int i = (0); i < (n); ++i)
    if (!bio[i]) dfs(i);
  int ptr = 0;
  for (int qhi = (0); qhi < (n); ++qhi) {
    from[qhi] = max(from[qhi], ptr);
    ptr = from[qhi];
    psum[qhi] += qhi - ptr + 1;
    for (int id = 0; id < lb[qhi].size(); ++id) {
      int qlo = lb[qhi][id];
      int lo = 0, hi = qhi, j = 0;
      while (lo <= hi) {
        int mid = (lo + hi) / 2;
        if (from[mid] <= qlo) {
          j = mid;
          lo = mid + 1;
        } else {
          hi = mid - 1;
        }
      }
      long long tot = 0;
      long long k = j - qlo + 1;
      tot += k * (k + 1) / 2;
      if (j + 1 <= qhi) tot += sum(j + 1, qhi);
      sol[idx[qhi][id]] = tot;
    }
    psum[qhi + 1] += psum[qhi];
  }
  for (int i = (0); i < (q); ++i) cout << sol[i] << endl;
  return 0;
}
