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;
long long int vis[300001], par[300001], dep[300001], beg[300001], A[300001];
vector<long long int> v[300001];
vector<pair<long long int, long long int> > cycles;
void dfs(long long int x) {
  vis[x] = 1;
  for (long long int y : v[x]) {
    if (y == par[x]) continue;
    if (vis[y]) {
      cycles.push_back(make_pair(x, y));
      continue;
    }
    dep[y] = dep[x] + 1;
    par[y] = x;
    dfs(y);
  }
}
void make_cycles(long long int n) {
  for (long long int i = 1; i <= n; ++i) beg[i] = 1;
  for (long long int i = 0; i < cycles.size(); ++i) {
    long long int x = cycles[i].first;
    long long int y = cycles[i].second;
    long long int maxv = max(x, y), minv = min(x, y);
    while (x != y) {
      if (dep[x] >= dep[y]) {
        x = par[x];
      } else {
        y = par[y];
      }
      maxv = max(maxv, max(x, y));
      minv = min(minv, min(x, y));
    }
    beg[maxv] = minv + 1;
  }
  for (long long int i = 1; i <= n; ++i) {
    beg[i] = max(beg[i], beg[i - 1]);
    A[i] = A[i - 1] + i - beg[i] + 1;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  long long int n, m, i, j, k, x, y, z, a, b, c;
  cin >> n >> m;
  for (i = 0; i < m; ++i) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (i = 1; i <= n; ++i) {
    if (!vis[i]) dfs(i);
  }
  make_cycles(n);
  cin >> m;
  while (m--) {
    cin >> x >> y;
    long long int start = x;
    long long int end = y;
    long long int ans = y + 1;
    while (start <= end) {
      long long int mid = (start + end) >> 1;
      if (beg[mid] >= x) {
        ans = mid;
        end = mid - 1;
      } else {
        start = mid + 1;
      }
    }
    long long int val = A[y] - A[ans - 1] + (ans - x) * (ans - x + 1) / 2;
    cout << val << endl;
  }
  return 0;
}
