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> G[303000];
int n, m, T, x, y, l, r, fa[303000], dep[303000], nex[303000];
long long val[303000];
void doit(int x, int y) {
  if (dep[x] < dep[y]) return;
  int mx = max(x, y), mn = min(x, y);
  for (; x - y; mx = max(x, mx), mn = min(x, mn)) x = fa[x];
  nex[mn] = min(nex[mn], mx - 1);
}
void dfs(int x, int fff) {
  fa[x] = fff, dep[x] = dep[fff] + 1;
  for (auto y : G[x])
    if (y - fff) {
      if (dep[y])
        doit(x, y);
      else
        dfs(y, x);
    }
}
int main() {
  for (cin >> n >> m; m--;) {
    scanf("%d%d", &x, &y), G[x].push_back(y), G[y].push_back(x);
  }
  for (int i = 1; i <= n; ++i) nex[i] = n;
  for (int i = 1; i <= n; ++i)
    if (!dep[i]) dfs(i, 0);
  for (int i = n - 1; i; --i) nex[i] = min(nex[i], nex[i + 1]);
  for (int i = 1; i <= n; ++i) val[i] = val[i - 1] + nex[i] - i + 1;
  for (cin >> T; T--;) {
    scanf("%d%d", &l, &r);
    x = lower_bound(nex + 1, nex + n + 1, r) - nex;
    x = max(x - 1, l - 1);
    printf("%I64d\n", val[x] - val[l - 1] + (r - x) * (r - x + 1ll) / 2);
  }
}
