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 MAX = 3e5 + 2;
vector<int> arr[MAX];
int vis[MAX], stk[MAX];
long long rgt[MAX], prf[MAX];
void dfs(int v, int p, int step) {
  vis[v] = 1;
  stk[step] = v;
  for (auto to : arr[v])
    if (!vis[to])
      dfs(to, v, step + 1);
    else if (to != p && vis[to] == 1) {
      int mx = v, mn = v;
      for (int i = step - 1; stk[i + 1] != to; --i) {
        mx = max(mx, stk[i]);
        mn = min(mn, stk[i]);
      }
      rgt[mn] = mx;
    }
  vis[v] = 2;
}
int main() {
  ios::sync_with_stdio(0);
  int n;
  long long m;
  cin >> n >> m;
  while (m--) {
    int from, to;
    cin >> from >> to;
    arr[from].push_back(to);
    arr[to].push_back(from);
  }
  for (int v = 1; v <= n + 1; ++v) rgt[v] = n + 1;
  for (int v = 1; v <= n; ++v)
    if (!vis[v]) dfs(v, 0, 1);
  for (int v = n; v >= 1; --v) rgt[v] = min(rgt[v], rgt[v + 1]);
  prf[1] = rgt[1] - 1;
  for (int i = 2; i <= n; ++i) prf[i] = rgt[i] - i + prf[i - 1];
  int q;
  cin >> q;
  while (q--) {
    long long from, to;
    cin >> from >> to;
    long long l = from - 1, r = to;
    while (l < r - 1) {
      m = (l + r) >> 1LL;
      if (rgt[m] <= to)
        l = m;
      else
        r = m;
    }
    cout << prf[l] - prf[from - 1] + (to - l) * (to - l + 1) / 2LL << endl;
  }
  return 0;
}
