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;
vector<int> way[300005];
vector<pair<int, int>> qr[300005];
int vs[300005];
int vs_tmp[300005];
int cnt[300005];
long long res[300005];
int lst[300005];
int idx;
int ban[300005];
void dfs(int pos, int from) {
  if (vs[pos]) return;
  vs[pos] = 1;
  vs_tmp[pos] = 1;
  lst[idx++] = pos;
  for (int v : way[pos]) {
    if (v == from) continue;
    if (vs_tmp[v]) {
      int mn = v, mx = v;
      for (int k = idx - 1; lst[k] != v; k--) {
        mn = min(mn, lst[k]);
        mx = max(mx, lst[k]);
      }
      ban[mx] = max(ban[mx], mn);
    }
    dfs(v, pos);
  }
  idx--;
  vs_tmp[pos] = 0;
  return;
}
long long t[2000000];
int lazy[2000000];
void add(int node, int l, int r, int targ_l, int targ_r) {
  if (r < targ_l) return;
  if (l > targ_r) return;
  if (targ_l <= l && r <= targ_r) {
    t[node] += r - l + 1;
    lazy[node]++;
    return;
  }
  int mid = (l + r) >> 1;
  lazy[node << 1] += lazy[node];
  lazy[node << 1 | 1] += lazy[node];
  t[node << 1] += 1LL * lazy[node] * (mid - l + 1);
  t[node << 1 | 1] += 1LL * lazy[node] * (r - mid);
  lazy[node] = 0;
  add(node << 1, l, mid, targ_l, targ_r);
  add(node << 1 | 1, mid + 1, r, targ_l, targ_r);
  t[node] = t[node << 1] + t[node << 1 | 1];
  return;
}
long long sum(int node, int l, int r, int targ_l, int targ_r) {
  if (r < targ_l) return 0;
  if (l > targ_r) return 0;
  if (targ_l <= l && r <= targ_r) return t[node];
  int mid = (l + r) >> 1;
  lazy[node << 1] += lazy[node];
  lazy[node << 1 | 1] += lazy[node];
  t[node << 1] += 1LL * lazy[node] * (mid - l + 1);
  t[node << 1 | 1] += 1LL * lazy[node] * (r - mid);
  lazy[node] = 0;
  long long res = 0;
  res += sum(node << 1, l, mid, targ_l, targ_r);
  res += sum(node << 1 | 1, mid + 1, r, targ_l, targ_r);
  t[node] = t[node << 1] + t[node << 1 | 1];
  return res;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    way[u].push_back(v);
    way[v].push_back(u);
  }
  for (int i = 1; i <= n; i++)
    if (!vs[i]) dfs(i, i);
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    qr[r].push_back({l, i});
  }
  int mx = 0;
  for (int i = 1; i <= n; i++) {
    mx = max(ban[i], mx);
    if (mx + 1 <= i) {
      add(1, 1, n, mx + 1, i);
    }
    for (auto p : qr[i]) {
      int l = p.first;
      int idx = p.second;
      res[idx] = sum(1, 1, n, l, i);
    }
  }
  for (int i = 1; i <= q; i++) printf("%lld\n", res[i]);
  return 0;
}
