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;
#pragma GCC optimize("O3")
#pragma GCC optimize("O2")
const long long arr = 500000;
vector<long long> vec[arr];
long long mx[arr], use[arr], sum[arr], used[arr];
;
vector<long long> visited;
void dfs(long long u, long long pred) {
  use[u] = 1;
  used[u] = 1;
  visited.push_back(u);
  for (auto i : vec[u]) {
    if (i == pred) continue;
    if (use[i]) {
      long long mxx = u;
      long long mnx = u;
      long long last = visited.size() - 1;
      while (visited[last] != i) {
        last--;
        mxx = max(mxx, visited[last]);
        mnx = min(mnx, visited[last]);
      }
      mx[mnx] = mxx;
    } else {
      if (!used[i]) dfs(i, u);
    }
  }
  use[u] = 0;
  visited.pop_back();
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, m;
  cin >> n >> m;
  for (long long i = 1; i <= m; i++) {
    long long a, b;
    cin >> a >> b;
    vec[a].push_back(b);
    vec[b].push_back(a);
  }
  for (long long i = 1; i <= n; i++) mx[i] = n + 1;
  for (long long i = 1; i <= n; i++)
    if (!used[i]) {
      dfs(i, 0);
    }
  for (long long i = n - 1; i >= 1; i--) mx[i] = min(mx[i], mx[i + 1]);
  for (long long i = 1; i <= n; i++) sum[i] = mx[i] + sum[i - 1];
  cin >> m;
  while (m--) {
    long long l, r;
    cin >> l >> r;
    long long res = 0;
    if (mx[l] > r + 1) {
      res += (r + 1) * (r - l + 1);
    } else {
      long long left = l;
      long long right = r;
      while (right - left > 1) {
        long long d = (right + left) / 2;
        if (mx[d] <= r + 1)
          left = d;
        else
          right = d;
      }
      if (mx[right] <= r + 1) {
        left = right;
      }
      res += sum[left] - sum[l - 1];
      res += (r - left) * (r + 1);
    }
    res -= (r + l) * (r - l + 1) / 2;
    cout << res << '\n';
  }
}
