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;
long long n, m, u, v, q;
bool come[330000];
vector<long long> edge[330000];
long long parent[330000];
long long group[330000];
long long g_min[330000];
long long g_max[330000];
long long limit[330000];
long long l_sum[330000];
long long dep[330000];
void dfs(long long x, long long last = -1) {
  come[x] = true;
  for (auto to : edge[x]) {
    if (come[to]) {
      if (dep[to] < dep[x] - 1) {
        for (long long now = x;; now = parent[now]) {
          group[now] = x;
          if (now == to) break;
        }
      }
      continue;
    }
    parent[to] = x;
    dep[to] = dep[x] + 1;
    dfs(to, x);
  }
}
void calc() {
  for (long long i = 1; i <= n; i++) g_min[i] = n;
  for (long long i = 1; i <= n; i++) {
    if (!come[i]) dfs(i);
    g_min[group[i]] = min(g_min[group[i]], i);
    g_max[group[i]] = max(g_max[group[i]], i);
  }
  limit[0] = 1;
  set<long long> groups = {group[1]};
  for (long long i = 1; i <= n; i++) {
    limit[i] = limit[i - 1];
    long long &l = limit[i];
    while (true) {
      if (l == n ||
          (g_max[group[l + 1]] == l + 1 && groups.count(group[l + 1])))
        break;
      l++;
      if (g_min[group[l]] == l) groups.insert(group[l]);
    }
    l_sum[i] = l_sum[i - 1] + (l - i + 1);
    if (g_min[group[i]] == i) groups.erase(group[i]);
  }
}
void solve() {
  long long l, r;
  scanf("%lld%lld", &l, &r);
  long long bottom = l;
  long long top = r;
  if (limit[l] >= r) {
    top = l;
    bottom = l - 1;
  }
  while (top - bottom > 1) {
    long long mid = (bottom + top) / 2;
    if (limit[mid] >= r)
      top = mid;
    else
      bottom = mid;
  }
  long long ans = (r - top + 1) * (r - top + 2) / 2;
  ans += l_sum[bottom] - l_sum[l - 1];
  printf("%lld\n", ans);
}
int main() {
  scanf("%lld%lld", &n, &m);
  for (long long i = 1; i <= n; i++) group[i] = i;
  for (long long i = 0; i < m; i++) {
    scanf("%lld%lld", &u, &v);
    edge[u].push_back(v);
    edge[v].push_back(u);
  }
  calc();
  scanf("%lld", &q);
  for (long long i = 0; i < q; i++) solve();
  return 0;
}
