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 gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 300000;
int n, m;
vector<int> adj[MAXN];
vector<int> cycle[MAXN];
int limit[MAXN];
long long limitsum[MAXN + 1];
int state[MAXN];
int stck[MAXN], nstck;
void dfs(int at, int par) {
  state[at] = 1;
  stck[nstck++] = at;
  for (int i = (0); i < (((int)(adj[at]).size())); ++i) {
    int to = adj[at][i];
    if (to == par) continue;
    if (state[to] == 0) {
      dfs(to, at);
    } else if (state[to] == 1) {
      int mn = to, mx = to;
      for (int i = nstck - 1; stck[i] != to; --i)
        mn = min(mn, stck[i]), mx = max(mx, stck[i]);
      cycle[mn].push_back(mx);
    }
  }
  state[at] = 2;
  assert(stck[--nstck] == at);
}
void run() {
  scanf("%d%d", &n, &m);
  for (int i = (0); i < (m); ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a, --b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  memset(state, 0, sizeof(state));
  for (int i = (0); i < (n); ++i) cycle[i].clear();
  for (int i = (0); i < (n); ++i)
    if (state[i] == 0) dfs(i, -1);
  for (int i = n - 1; i >= 0; --i) {
    limit[i] = i == n - 1 ? n - 1 : limit[i + 1];
    for (int j = (0); j < (((int)(cycle[i]).size())); ++j)
      limit[i] = min(limit[i], cycle[i][j] - 1);
  }
  limitsum[0] = 0;
  for (int i = (0); i < (n); ++i)
    limitsum[i + 1] = limitsum[i] + (limit[i] - i + 1);
  int nq;
  scanf("%d", &nq);
  for (int qi = (0); qi < (nq); ++qi) {
    int ql, qr;
    scanf("%d%d", &ql, &qr);
    --ql, --qr;
    int l = ql - 1, r = qr + 1;
    while (l + 1 < r) {
      int m = l + (r - l) / 2;
      if (limit[m] > qr)
        r = m;
      else
        l = m;
    }
    int qm = r;
    long long ans = 0;
    int nsmall = qm - ql, nlarge = qr - qm + 1;
    ans += (long long)(limitsum[qm] - limitsum[ql]);
    ans += (long long)nlarge * (nlarge + 1) / 2;
    printf("%lld\n", ans);
  }
}
int main() {
  run();
  return 0;
}
