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;
vector<int> g[300100];
int n, m, q;
int vs[300100], fa[300100];
int val[300100];
void dfs(int u, int pre) {
  vs[u] = 1;
  fa[u] = pre;
  for (auto v : g[u]) {
    if (v == pre || vs[v] == 2) continue;
    if (!vs[v])
      dfs(v, u);
    else {
      int mn = v, mx = v, now = u;
      while (now != v) {
        mn = min(mn, now);
        mx = max(mx, now);
        now = fa[now];
      }
      val[mn] = min(val[mn], mx - 1);
    }
  }
  vs[u] = 2;
}
long long sum[300100];
inline long long cal(long long x) { return x * (x + 1) / 2; }
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; ++i) {
    val[i] = m;
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 1; i <= n; ++i) val[i] = n;
  for (int i = 1; i <= n; ++i)
    if (!vs[i]) dfs(i, -1);
  for (int i = n - 1; i >= 1; --i) val[i] = min(val[i], val[i + 1]);
  for (int i = 1; i <= n; ++i) sum[i] = sum[i - 1] + val[i] - i + 1;
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    int lb = l - 1, ub = r;
    while (ub > lb + 1) {
      int mid = (lb + ub) >> 1;
      if (val[mid] >= r)
        ub = mid;
      else
        lb = mid;
    }
    long long ans = sum[ub - 1] - sum[l - 1];
    ans += cal(r - ub + 1);
    printf("%I64d\n", ans);
  }
  return 0;
}
