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 = 300000;
int n, m;
vector<int> adj[N];
bool visited[N];
int parent[N], depth[N], mx[N + 1];
long long psum[N + 1];
void dfs(int here) {
  visited[here] = true;
  for (int next : adj[here]) {
    if (next == parent[here]) continue;
    if (visited[next]) {
      if (depth[here] < depth[next]) continue;
      int p = here, cyclemin = here, cyclemax = here;
      while (p != next) {
        p = parent[p];
        if (cyclemin > p) cyclemin = p;
        if (cyclemax < p) cyclemax = p;
      }
      mx[cyclemin + 1] = cyclemax;
    } else {
      parent[next] = here;
      depth[next] = depth[here] + 1;
      dfs(next);
    }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0, a, b; i < m; ++i) {
    scanf("%d%d", &a, &b);
    --a;
    --b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  for (int i = 1; i <= n; ++i) mx[i] = n;
  for (int i = 0; i < n; ++i) {
    if (visited[i]) continue;
    parent[i] = -1;
    dfs(i);
  }
  for (int i = n - 1; i > 0; --i) mx[i] = min(mx[i], mx[i + 1]);
  for (int i = 0; i < n; ++i) psum[i + 1] = psum[i] + mx[i + 1] - i;
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    int lo = l, hi = r;
    while (lo < hi) {
      int mid = (lo + hi) / 2;
      if (mx[mid] >= r)
        hi = mid;
      else
        lo = mid + 1;
    }
    long long ans = 0;
    ans += (long long)(r - lo + 1) * (r - lo + 2) / 2;
    ans += psum[lo - 1] - psum[l - 1];
    printf("%lld\n", ans);
  }
  return 0;
}
