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 long long one = 1;
const long long INF = 1e9 + 100;
const long long MOD = 1e9 + 7;
const long long N = 3e5 + 100;
const long long nul = 0;
const long long SIZE = 1e6 * 3;
double getTime() { return clock() / (double)CLOCKS_PER_SEC; };
vector<long long> gr[N];
long long DELETE[N];
long long p[N];
long long rev[N];
long long tin[N];
long long t = 0;
void dfs(long long v) {
  tin[v] = t++;
  for (auto u : gr[v]) {
    if (u == p[v]) continue;
    if (p[u]) {
      if (tin[u] < tin[v]) rev[v] = u;
      continue;
    }
    p[u] = v;
    dfs(u);
  }
}
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  ;
  long long n, m;
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    long long v, u;
    cin >> v >> u;
    gr[v].push_back(u);
    gr[u].push_back(v);
  }
  for (long long i = 1; i <= n; i++) {
    if (!p[i]) {
      p[i] = -1;
      dfs(i);
    }
  }
  for (long long i = 1; i <= n; i++) {
    if (rev[i]) {
      long long v = rev[i];
      long long u = i;
      vector<long long> cycle = {v};
      while (u != v) {
        cycle.push_back(u);
        u = p[u];
      }
      long long mn = *min_element((cycle).begin(), (cycle).end());
      long long mx = *max_element((cycle).begin(), (cycle).end());
      DELETE[mx] = mn;
    }
  }
  vector<long long> prefix(n + 5, 0);
  vector<long long> time_del(n + 5, n + 5);
  time_del[0] = 1;
  for (long long i = 1, j = 0; i <= n; i++) {
    while (j < DELETE[i]) {
      j++;
      time_del[j] = i;
    }
    prefix[i] = prefix[i - 1] + i - j;
  }
  long long q;
  cin >> q;
  while (q--) {
    long long l, r;
    cin >> l >> r;
    long long l2 = min(time_del[l - 1] - 1, r);
    cout << prefix[r] - prefix[l2] + (l2 - l + 1) * (l2 - l + 2) / 2 << "\n";
  }
}
