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>
const int MAXN = 300005;
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int n, m, qn, cnt, vis[MAXN], pa[MAXN];
long long lb[MAXN], arith[MAXN], openl[MAXN], pref[MAXN];
vector<vector<int>> cyc;
vector<int> G[MAXN];
void dfs(int v, int p) {
  vis[v] = 1;
  pa[v] = p;
  for (int to : G[v]) {
    if (vis[to] == 0) {
      dfs(to, v);
    } else if (vis[to] == 1 && to != p) {
      int cur = v;
      ++cnt;
      cyc.push_back(vector<int>());
      while (cur != to) {
        cyc.back().push_back(cur);
        cur = pa[cur];
      }
      cyc.back().push_back(to);
    }
  }
  vis[v] = 2;
}
void solve() {
  cin >> n >> m;
  cyc.push_back(vector<int>());
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      dfs(i, i);
    }
  }
  for (int i = 1; i <= cnt; i++) {
    sort((cyc[i]).begin(), (cyc[i]).end());
    lb[cyc[i].back()] = max(lb[cyc[i].back()], (long long)cyc[i][0]);
  }
  long long curl = 0;
  for (int i = 1; i <= n; i++) {
    curl = max(curl, lb[i]);
    openl[i] = curl;
  }
  for (int i = 1; i <= n; i++) {
    pref[i] = pref[i - 1] + openl[i];
    arith[i] = arith[i - 1] + i;
  }
  cin >> qn;
  for (int q = 1; q <= qn; q++) {
    long long l, r;
    cin >> l >> r;
    long long st = lower_bound(openl + 1, openl + 1 + n, l) - openl,
              ans = arith[r] - arith[l - 1];
    st = max(st, l);
    st = min(st, r + 1);
    if (l < st) {
      ans -= (l - 1) * (st - l);
    }
    if (st <= r) {
      ans -= pref[r] - pref[st - 1];
    }
    cout << ans << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int tc = 1;
  for (int i = 1; i <= tc; i++) {
    solve();
  }
  return 0;
}
