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 N = 300010;
int n, m, q, x, y, l, r, mid, cnt, dfn[N], low[N], s[N], R[N], fa[N], t, v,
    st[N];
int p, head[N], to[N << 1], nxt[N << 1];
long long sum[N], ans;
inline void add(int x, int y) {
  to[++p] = y;
  nxt[p] = head[x];
  head[x] = p;
  to[++p] = x;
  nxt[p] = head[y];
  head[y] = p;
}
void tarjan(int x, int fa) {
  s[++s[0]] = x;
  low[x] = dfn[x] = ++cnt;
  for (int i = head[x]; i; i = nxt[i])
    if (to[i] != fa) {
      if (dfn[to[i]])
        low[x] = min(low[x], low[to[i]]);
      else {
        tarjan(to[i], x);
        low[x] = min(low[x], low[to[i]]);
        if (low[to[i]] >= dfn[x]) {
          t = 0;
          l = r = x;
          v = 0;
          do {
            t = s[s[0]--];
            v++;
            l = min(l, t);
            r = max(r, t);
          } while (t != to[i]);
          if (v == 1) continue;
          R[l] = min(R[l], r - 1);
        }
      }
    }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) scanf("%d%d", &x, &y), add(x, y);
  for (int i = 1; i <= n; i++) R[i] = n;
  for (int i = 1; i <= n; i++)
    if (!dfn[i]) tarjan(i, 0);
  for (int i = n - 1; i; i--) R[i] = min(R[i], R[i + 1]);
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + (long long)R[i];
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &x, &y);
    if (R[x] >= y)
      printf("%lld\n", 1ll * (y - x + 1) * (y - x + 2) / 2);
    else {
      l = x;
      r = y;
      while (l < r) {
        mid = (l + r) >> 1;
        if (R[mid] >= y)
          r = mid;
        else
          l = mid + 1;
      }
      printf("%lld\n", sum[l - 1] - sum[x - 1] + 1ll * (y - l + 1) * y -
                           1ll * (y - x + 1) * (x + y - 2) / 2);
    }
  }
  return 0;
}
