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;
int n, m, num, head[300010], from, to, p[300010];
long long sum[300010];
int v[300010];
int read() {
  int x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = getchar();
  }
  return x;
}
struct xx {
  int next, to;
} way[600010];
void add(int from, int to) {
  way[++num].next = head[from];
  way[num].to = to;
  head[from] = num;
}
stack<int> q;
void dfs(int x, int fa) {
  v[x] = 1;
  q.push(x);
  for (int i = head[x]; i; i = way[i].next) {
    int y = way[i].to;
    if (y != fa) {
      if (v[y] == 2) continue;
      if (v[y] == 0)
        dfs(y, x);
      else {
        int maxn = y;
        int minl = y;
        while (q.size()) {
          int w = q.top();
          q.pop();
          maxn = max(maxn, w);
          minl = min(minl, w);
          if (w == y) break;
        }
        p[maxn] = minl;
      }
    }
  }
  v[x] = 2;
  if (q.size() && q.top() == x) q.pop();
}
int main() {
  n = read();
  m = read();
  for (int i = 1; i <= m; i++) {
    from = read();
    to = read();
    add(from, to);
    add(to, from);
  }
  for (int i = 1; i <= n; i++)
    if (v[i] == 0) dfs(i, 0);
  p[0] = 0;
  for (int i = 1; i <= n; i++)
    p[i] = max(p[i - 1], p[i]), sum[i] = sum[i - 1] + i - p[i];
  int q = read();
  while (q--) {
    int l, r;
    l = read();
    r = read();
    if (p[r] + 1 <= l) {
      printf("%lld\n", 1ll * (r - l + 2) * (r - l + 1) / 2);
      continue;
    }
    int ll = l, rr = r;
    while (ll < rr) {
      int mid = (ll + rr) >> 1;
      if (p[mid] + 1 <= l)
        ll = mid + 1;
      else
        rr = mid;
    }
    long long ans = sum[r] - sum[ll - 1] + 1ll * (ll - l + 1) * (ll - l) / 2;
    printf("%lld\n", ans);
  }
  return 0;
}
