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 long long p = 1000000007;
vector<pair<int, int> > cycles;
vector<int> graph[300002];
short visited[300002];
int low[300002];
int mm[300002];
stack<int> s;
void d_edge(int a, int b) { graph[a].push_back(b); }
void edge(int a, int b) {
  graph[a].push_back(b);
  graph[b].push_back(a);
}
void dfs(int from, int vert, int l) {
  visited[vert] = 1;
  low[vert] = l;
  for (auto nb : graph[vert]) {
    if (nb == from) continue;
    s.push(vert);
    if (visited[nb] == 1) {
      int mini = 1000000;
      int maxi = 0;
      while (1) {
        int top = s.top();
        low[top] = min(low[top], low[nb]);
        maxi = max(maxi, top);
        mini = min(mini, top);
        s.pop();
        if (top == nb) break;
      }
      mm[maxi] = max(mm[maxi], mini);
    }
    if (visited[nb] == 0) {
      dfs(vert, nb, l + 1);
    }
  }
  visited[vert] = 2;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n + 1; i++) mm[i] = -1;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    edge(a, b);
  }
  for (int i = 0; i < n; i++) {
    if (!visited[i + 1]) {
      dfs(-1, i + 1, 0);
    }
  }
  int last_b[n + 1];
  int i = 1, j = 1;
  while (i <= n) {
    while (mm[j] < i && j <= n) j++;
    last_b[i] = j;
    i++;
  }
  long long pr_sum[n + 1];
  pr_sum[0] = 0;
  for (int i = 1; i < n + 1; i++) {
    pr_sum[i] = pr_sum[i - 1] + last_b[i];
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    long long result = 0;
    long long a, b;
    cin >> a >> b;
    long long beg = a;
    long long end = b;
    while (beg < end) {
      int s = (beg + end) / 2;
      if (last_b[s] > b) {
        end = s;
      } else {
        beg = s + 1;
      }
    }
    result = (pr_sum[beg - 1] - pr_sum[a - 1]) -
             (((beg - 1) * beg) / 2 - ((a - 1) * a) / 2);
    result += ((b - beg + 1) * (b - beg + 2)) / 2;
    cout << result << endl;
  }
}
