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 N = 3000 * 100 + 10;
vector<int> adj[N];
vector<pair<int, int> > cy[N];
long long pr[N];
int h[N], num[N], vert[N], en[N], c, n;
bool mark[N];
void DFS(int v, int par) {
  mark[v] = 1;
  for (int u : adj[v]) {
    if (!mark[u]) {
      h[u] = h[v] + 1;
      DFS(u, v);
      for (auto k : cy[u])
        if (k.second <= h[v]) cy[v].push_back(k);
    } else if (h[u] < h[v] && u != par) {
      cy[v].push_back(make_pair(c, h[u]));
      vert[c] = h[v] - h[u] + 1;
      c++;
    }
  }
  return;
}
bool check(int v) {
  for (auto u : cy[v])
    if (num[u.first] + 1 == vert[u.first]) return 0;
  return 1;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int m, q, u, v;
  long long l, r;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> u >> v;
    adj[--u].push_back(--v);
    adj[v].push_back(u);
  }
  for (int i = 0; i < n; i++)
    if (!mark[i]) DFS(i, i);
  int pt = 0, pt2 = 0;
  while (pt < n) {
    if (pt2 < n && check(pt2)) {
      for (auto u : cy[pt2]) num[u.first]++;
      pt2++;
    } else {
      pr[pt] = pt2 - pt;
      en[pt] = pt2 - 1;
      for (auto u : cy[pt]) num[u.first]--;
      pt++;
    }
  }
  for (int i = 1; i < n; i++) pr[i] += pr[i - 1];
  cin >> q;
  for (int i = 0; i < q; i++) {
    cin >> l >> r;
    r--, l--;
    long long k = lower_bound(en, en + n, r + 1) - en - 1;
    k = min(k, r);
    long long ans = 0;
    if (k >= l) {
      ans += pr[k];
      if (l) ans -= pr[l - 1];
    } else
      k = l - 1;
    ans += (1 + r - k) * (r - k) / 2;
    cout << ans << endl;
  }
  return 0;
}
