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;
unordered_map<long long, unordered_set<long long> > adj;
vector<long long> vis1, lft, par;
long long c = 1;
void mark(long long loc, long long till) {
  long long mx = max(loc, till);
  long long mn = min(loc, till);
  while (loc != till) {
    loc = par[loc];
    mx = max(loc, mx);
    mn = min(loc, mn);
  }
  lft[mx] = mn + 1;
}
void dfs1(long long loc) {
  if (vis1[loc]) return;
  vis1[loc] = c++;
  for (auto &i : adj[loc]) {
    adj[i].erase(loc);
    if (!vis1[i]) {
      par[i] = loc;
      dfs1(i);
    } else if (vis1[i] < vis1[loc])
      mark(loc, i);
  }
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, m, x, y, q;
  cin >> n >> m;
  vis1.assign(n + 1, 0);
  lft.assign(n + 1, 1);
  par.assign(n + 1, 0);
  for (long long i = 0; i < m; i++) {
    cin >> x >> y;
    adj[x].insert(y);
    adj[y].insert(x);
  }
  for (long long i = 1; i <= n; i++) {
    dfs1(i);
  }
  vector<long long> lsum(n + 1, 0);
  for (long long i = 1; i <= n; i++) {
    lft[i] = max(lft[i], lft[i - 1]);
    lsum[i] = (i - lft[i] + 1) + lsum[i - 1];
  }
  cin >> q;
  for (long long i = 0; i < q; i++) {
    cin >> x >> y;
    long long loc = lower_bound(lft.begin() + 1, lft.end(), x) - lft.begin();
    loc = max(x, loc);
    long long ans = 0;
    if (loc > y) {
      long long len = (y - x + 1);
      ans = len * (len + 1) / 2;
    } else {
      ans = lsum[y] - lsum[loc - 1];
      long long len = (loc - x);
      ans += len * (len + 1) / 2;
    }
    cout << ans << '\n';
  }
  return 0;
}
