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 = 300005;
struct Edg {
  int lst, ed;
} e[N << 1];
int ne = 0, nte[N];
int tot = 0, top = 0, dfn[N], low[N], sta[N];
int d[N], a[N];
long long s[N];
int rd() {
  int x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
  return x;
}
void NewEdg(int u, int v) {
  ne++;
  e[ne].lst = nte[u];
  e[ne].ed = v;
  nte[u] = ne;
}
void dfs(int x) {
  dfn[x] = low[x] = ++tot;
  sta[++top] = x;
  for (int i = nte[x]; i; i = e[i].lst)
    if (!dfn[e[i].ed]) {
      dfs(e[i].ed);
      low[x] = min(low[x], low[e[i].ed]);
      if (low[e[i].ed] >= dfn[x]) {
        int L = x, R = x, siz = 1;
        while (sta[top + 1] != e[i].ed) {
          R = max(R, sta[top]);
          L = min(L, sta[top]);
          top--, siz++;
        }
        if (siz > 2) d[L] = min(d[L], R);
      }
    } else
      low[x] = min(low[x], dfn[e[i].ed]);
}
int main() {
  int n = rd(), m = rd();
  for (int i = 1; i <= m; i++) {
    int x = rd(), y = rd();
    NewEdg(x, y), NewEdg(y, x);
  }
  for (int i = 1; i <= n; i++) d[i] = n + 1;
  for (int i = 1; i <= n; i++)
    if (!dfn[i]) dfs(i);
  a[n] = 1;
  for (int i = n - 1; i >= 1; i--) a[i] = min(d[i] - i, a[i + 1] + 1);
  for (int i = 1; i <= n; i++) s[i] = s[i - 1] + a[i];
  int q = rd();
  for (int i = 1; i <= q; i++) {
    int L = rd(), R = rd();
    int l = L - 1, r = R + 1;
    while (l + 1 < r) {
      int mid = (l + r) >> 1;
      if (mid + a[mid] - 1 > R)
        r = mid;
      else
        l = mid;
    }
    printf("%lld\n", s[l] - s[L - 1] + (long long)(1 + R - l) * (R - l) / 2);
  }
  return 0;
}
