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<int> v[300010];
int f[300010], pre[300010], st[300010];
bool vis[300010];
long long sum1[300010], sum2[300010];
int n, m, T, top;
int find(int i) {
  if (f[i] == i) return i;
  if (!vis[i]) st[++top] = i, vis[i] = 1;
  return f[i] = find(f[i]);
}
void solve(int l, int r, int L, int R) {
  if (l > r) return;
  if (L == R) {
    for (int i = l; i <= r; i++) pre[i] = L;
    return;
  }
  int mid = (L + R) >> 1;
  int ans = r + 1;
  bool w = 1;
  for (int i = mid; i <= r; i++) {
    int j = lower_bound(v[i].begin(), v[i].end(), mid) - v[i].begin();
    for (; j < v[i].size(); j++) {
      int f1 = find(v[i][j]), f2 = find(i);
      if (f1 != f2) {
        f[f1] = f2;
        if (!vis[f1]) st[++top] = f1, vis[f1] = 1;
      } else {
        w = 0;
        break;
      }
    }
    if (!w) {
      ans = i;
      break;
    }
  }
  while (top) f[st[top]] = st[top], vis[st[top]] = 0, top--;
  solve(l, ans - 1, L, mid);
  solve(ans, r, mid + 1, R);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    v[max(x, y)].push_back(min(x, y));
  }
  for (int i = 1; i <= n; i++) sort(v[i].begin(), v[i].end());
  for (int i = 1; i <= n; i++) f[i] = i;
  solve(1, n, 1, n);
  for (int i = 1; i <= n; i++)
    sum1[i] = sum1[i - 1] + i, sum2[i] = sum2[i - 1] + (i - pre[i] + 1);
  scanf("%d", &T);
  while (T--) {
    int l, r;
    scanf("%d%d", &l, &r);
    long long ans = 0;
    int now = upper_bound(pre + l, pre + r + 1, l) - pre;
    ans = sum1[now - 1] - sum1[l - 1] - 1ll * (now - l) * (l - 1) + sum2[r] -
          sum2[now - 1];
    printf("%I64d\n", ans);
  }
  return 0;
}
