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 = 303000;
int n, m, dfn[maxn], dfs_clock;
vector<int> G[maxn], pa;
vector<pair<int, int> > cyc;
void dfs(int u, int fa) {
  dfn[u] = ++dfs_clock;
  pa.push_back(u);
  for (auto v : G[u])
    if (v != fa) {
      if (dfn[v]) {
        if (dfn[v] < dfn[u]) {
          pair<int, int> cc = {v, v};
          auto cur = pa;
          while (cur.back() != v) {
            cc.first = min(cc.first, cur.back());
            cc.second = max(cc.second, cur.back());
            cur.pop_back();
          }
          cyc.push_back(cc);
        }
      } else
        dfs(v, u);
    }
  pa.pop_back();
}
int rb[maxn], cnt[maxn];
long long pre[maxn];
int bs(int lpos, int rpos) {
  int l = lpos, r = rpos;
  while (l < r) {
    int mid = (l + r) >> 1;
    if (rb[mid] < rpos)
      l = mid + 1;
    else
      r = mid;
  }
  return l;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; ++i) {
    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)
    if (!dfn[i]) dfs(i, -1);
  for (int i = 1; i <= n; ++i) rb[i] = n;
  for (auto [l, r] : cyc) rb[l] = min(rb[l], r - 1);
  for (int i = n - 1; i > 0; --i) {
    rb[i] = min(rb[i], rb[i + 1]);
    cnt[i] = rb[i] - i + 1;
  }
  for (int i = 1; i <= n; ++i) pre[i] = pre[i - 1] + cnt[i];
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    int o = bs(l, r);
    printf("%lld\n",
           pre[o - 1] - pre[l - 1] + 1LL * (r - o + 2) * (r - o + 1) / 2);
  }
  return 0;
}
