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 N = 310000;
vector<int> V[N], adj[N];
int as[N], vis[N], par[N], T;
long long sum[N];
void dfs(int u, int p) {
  vis[u] = ++T, par[u] = p;
  for (int v : adj[u]) {
    if (v == p) continue;
    if (!vis[v])
      dfs(v, u);
    else if (vis[v] < vis[u]) {
      int x = u;
      int L = 1e6, R = 1;
      while (x != v) {
        L = min(L, x);
        R = max(R, x);
        x = par[x];
      }
      assert(x == v);
      L = min(L, v);
      R = max(R, v);
      V[L].push_back(R);
    }
  }
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i, 0);
  as[n + 1] = n + 1;
  for (int i = n; i > 0; i--) {
    as[i] = as[i + 1];
    for (int x : V[i]) as[i] = min(as[i], x);
  }
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + as[i] - i;
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    int p = upper_bound(as + l, as + n + 1, r) - as;
    printf("%lld\n",
           sum[p - 1] - sum[l - 1] + (r - p + 1) * 1LL * (r - p + 2) / 2);
  }
  return 0;
}
