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 + 5;
vector<int> g[maxn];
int n, m, p, a, b, vis[maxn];
long long ll, rr, l[maxn];
long long sum[maxn];
stack<int> s;
void dfs(int u, int f) {
  s.push(u);
  vis[u] = 1;
  int v;
  for (int i = 0; i < g[u].size(); i++) {
    v = g[u][i];
    if (v != f) {
      if (!vis[v])
        dfs(v, u);
      else if (vis[v] == 1) {
        int mn = v, mx = v;
        while (!s.empty()) {
          int cur = s.top();
          s.pop();
          mn = min(mn, cur);
          mx = max(mx, cur);
          if (cur == v) break;
        }
        l[mx] = mn;
      }
    }
  }
  if (!s.empty() && s.top() == u) s.pop();
  vis[u] = 2;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i, 0);
  for (int i = 1; i <= n; i++) {
    l[i] = max(l[i], l[i - 1]);
    sum[i] = sum[i - 1] + (1LL) * (i - l[i]);
  }
  scanf("%d", &p);
  for (int i = 0; i < p; i++) {
    scanf("%I64d%I64d", &ll, &rr);
    long long k = lower_bound(l + ll, l + rr + 1, ll) - l;
    long long ret = (k - ll) * (k - ll + 1) / 2 + sum[rr] - sum[k - 1];
    printf("%I64d\n", ret);
  }
}
