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>
std::vector<int> v[1000007];
bool odw[1000007];
bool jest[1000007];
std::vector<int> ter;
std::vector<int> cykle[1000007];
int dp[1000007];
long long pref[1000007];
void dfs(int a, int oj) {
  odw[a] = true;
  ter.push_back(a);
  jest[a] = true;
  for (auto it : v[a]) {
    if (!odw[it]) {
      dfs(it, a);
    } else if (jest[it] && it != oj) {
      int mini = it, maxi = it;
      auto w = ter.rbegin();
      while (*w != it) {
        mini = std::min(mini, *w);
        maxi = std::max(maxi, *w);
        w++;
      }
      cykle[maxi].push_back(mini);
    }
  }
  ter.pop_back();
  jest[a] = false;
}
int main() {
  int N, M;
  scanf("%d%d", &N, &M);
  int a, b;
  for (int i = 0; i < M; i++) {
    scanf("%d%d", &a, &b);
    v[a].push_back(b);
    v[b].push_back(a);
  }
  for (int i = 1; i <= N; i++)
    if (!odw[i]) dfs(i, -1);
  for (int i = 1; i <= N; i++) std::sort(cykle[i].begin(), cykle[i].end());
  int kon = 1;
  for (int i = 1; i <= N; i++) {
    while (kon <= N - 1 &&
           (cykle[kon + 1].empty() || *cykle[kon + 1].begin() < i)) {
      if (!cykle[kon + 1].empty())
        cykle[kon + 1].pop_back();
      else
        kon++;
    }
    dp[i] = kon;
    pref[i] = pref[i - 1] + dp[i] - i + 1;
  }
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    scanf("%d%d", &a, &b);
    int sr = int(std::lower_bound(dp + a, dp + b + 1, b) - dp);
    long long ile = b - sr + 1;
    long long wyn = ile * (ile + 1) / 2 + pref[sr - 1] - pref[a - 1];
    printf("%lld \n", wyn);
  }
  return 0;
}
