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 MAXN = 3e5 + 10;
int n, m, q;
bool mark[MAXN];
int mn[MAXN], h[MAXN], par[MAXN];
long long ps[MAXN];
vector<int> adj[MAXN];
void dfs(int u, int p = -1) {
  mark[u] = true;
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (mark[v] && p != v && h[u] > h[v]) {
      int mx = u, mnn = u, cur = u;
      while (cur != v) {
        cur = par[cur];
        mx = max(mx, cur), mnn = min(mnn, cur);
      }
      mn[mnn] = min(mn[mnn], mx);
    } else if (!mark[v]) {
      par[v] = u;
      h[v] = h[u] + 1;
      dfs(v, u);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b, a--, b--;
    adj[a].push_back(b), adj[b].push_back(a);
  }
  fill(mn, mn + n, n);
  for (int i = 0; i < n; i++) {
    if (!mark[i]) dfs(i);
  }
  for (int i = n - 2; i >= 0; i--) mn[i] = min(mn[i + 1], mn[i]);
  ps[0] = mn[0];
  for (int i = 1; i < n; i++) ps[i] = ps[i - 1] + mn[i] - i;
  cin >> q;
  while (q) {
    q--;
    int l, r;
    cin >> l >> r, l--, r--;
    int id = lower_bound(mn + l, mn + r + 1, r + 1) - mn;
    long long res = (id ? ps[id - 1] : 0) - (l ? ps[l - 1] : 0) +
                    1LL * (r - id + 1) * (r - id + 2) / 2;
    cout << res << endl;
  }
  return 0;
}
