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 = 3e5 + 77, Mod = 1e9 + 7;
int n, m, par[N], R[N], h[N], q, l, r;
long long ps[N];
vector<int> a[N];
bool M[N];
void dfs(int v) {
  M[v] = true;
  for (int u : a[v]) {
    if (M[u] && h[u] < h[v] && par[v] != u) {
      int x = v, Mx = v, Mn = v;
      while (x != u) {
        x = par[x];
        Mn = min(Mn, x);
        Mx = max(Mx, x);
      }
      R[Mn] = min(R[Mn], Mx);
    }
    if (M[u]) continue;
    par[u] = v;
    h[u] = h[v] + 1;
    dfs(u);
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) R[i] = n + 1;
  R[n + 1] = n + 2;
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    a[x].push_back(y);
    a[y].push_back(x);
  }
  for (int i = 1; i <= n; i++)
    if (!M[i]) dfs(i);
  for (int i = n - 1; i > 0; i--) R[i] = min(R[i], R[i + 1]);
  for (int i = 1; i <= n; i++) ps[i] = ps[i - 1] + R[i] - i;
  scanf("%d", &q);
  while (q--) {
    scanf("%d %d", &l, &r);
    int x = (lower_bound(R + l, R + r + 1, r + 1) - R);
    printf("%I64d\n",
           -ps[l - 1] + ps[x - 1] + ((r - x + 1) * 1ll * (r - x + 2) / 2));
  }
}
