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<long long> G[500009];
stack<long long> S;
long long vis[500009], p[500009], sum[500009];
long long cnt = 0;
void DFS(long long u, long long pa) {
  long long t;
  cnt++;
  vis[u] = cnt;
  S.push(u);
  for (auto v : G[u]) {
    if (v == pa) continue;
    if (!vis[v])
      DFS(v, u);
    else if (vis[v] < vis[u]) {
      long long mx = u, mn = u;
      while (!S.empty()) {
        t = S.top();
        S.pop();
        mx = max(mx, t);
        mn = min(mn, t);
        if (t == v) break;
      }
      p[mn] = mx;
    }
  }
}
int main() {
  long long N, M, i, j, k, a, b, l, r, L, R, Q;
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> M;
  for (i = 0; i < M; i++) {
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  for (i = 1; i <= N; i++) {
    p[i] = N + 1;
  }
  for (i = 1; i <= N; i++) {
    if (vis[i] == 0) DFS(i, 0);
  }
  sum[N] = 1;
  for (i = N - 1; i >= 1; i--) {
    p[i] = min(p[i], p[i + 1]);
    sum[i] = sum[i + 1] + p[i] - i;
  }
  cin >> Q;
  for (i = 0; i < Q; i++) {
    cin >> l >> r;
    L = l - 1;
    R = r;
    while (L + 1 < R) {
      long long mid = (L + R) / 2;
      if (p[mid] > r) {
        R = mid;
      } else {
        L = mid;
      }
    }
    cout << sum[l] - sum[R] + (r - R + 1) * (r - R + 2) / 2 << endl;
  }
}
