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 + 20;
vector<int> adj[maxn], rq[maxn];
int par[maxn], a[maxn], h[maxn];
long long sum[maxn];
bool visited[maxn];
void dfs(int v) {
  visited[v] = 1;
  for (auto u : adj[v])
    if (u != par[v]) {
      if (!visited[u]) {
        h[u] = h[v] + 1;
        par[u] = v;
        dfs(u);
      } else if (h[u] < h[v]) {
        int tmp = v, mn = u, mx = u;
        while (tmp != u) {
          mn = min(mn, tmp);
          mx = max(mx, tmp);
          tmp = par[tmp];
        }
        rq[mx].push_back(mn);
      }
    }
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    adj[--a].push_back(--b);
    adj[b].push_back(a);
  }
  for (int i = 0; i < n; i++)
    if (!visited[i]) {
      par[i] = -1;
      dfs(i);
    }
  int lst = -1;
  for (int i = 0; i < n; i++) {
    for (auto x : rq[i]) lst = max(lst, x);
    a[i] = lst;
  }
  for (int i = 0; i < n; i++) sum[i + 1] = sum[i] + (i - a[i]);
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    long long res = 0;
    long long k = lower_bound(a, a + n, l) - a;
    if (k < r) res += sum[r] - sum[max(k, (long long)l)];
    k = min(k, (long long)r);
    if (l <= k) res += (long long)(k - l) * (k - l + 1) / 2;
    printf("%I64d\n", res);
  }
}
