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 = 300009;
vector<int> neigh[maxn];
int mn[maxn];
int depth[maxn];
int st[maxn];
int st_sz = 0;
void dfs(int v, int p) {
  st[st_sz++] = v;
  for (int u : neigh[v]) {
    if (u == p) continue;
    if (depth[u] == -1) {
      depth[u] = depth[v] + 1;
      dfs(u, v);
    } else if (depth[u] < depth[v]) {
      int a = 0, b = maxn;
      for (int i = depth[v]; i >= depth[u]; i--) {
        a = max(a, st[i]);
        b = min(b, st[i]);
      }
      mn[b] = min(mn[b], a - 1);
    }
  }
  st_sz--;
}
long long pref_sum[maxn];
int low[maxn];
int to[maxn];
int main() {
  ios::sync_with_stdio(false);
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    neigh[u].push_back(v);
    neigh[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) {
    mn[i] = n;
    depth[i] = -1;
  }
  for (int v = 1; v <= n; v++) {
    if (depth[v] == -1) {
      depth[v] = 0;
      dfs(v, 0);
    }
  }
  for (int v = n - 1; v >= 1; v--) {
    mn[v] = min(mn[v], mn[v + 1]);
  }
  for (int v = 1, j = 1; v <= n; v++) {
    pref_sum[v] = pref_sum[v - 1] + mn[v] - v + 1;
    while (j <= v && mn[j] <= v) j++;
    to[v] = j;
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    long long ans = 0;
    int rr = max(to[r], l) - 1;
    ans =
        pref_sum[rr] - pref_sum[l - 1] + 1LL * (r - rr) * (r - rr + 1LL) / 2LL;
    printf("%I64d\n", ans);
  }
}
