#include <bits/stdc++.h>
using namespace std;
long long n, m, q;
long long a, b, tmp;
long long vis[300000 + 5];
vector<long long> path[300000 + 5];
stack<long long> cir;
long long p[300000 + 5], pre[300000 + 5];
void DFS(long long u, long long pa) {
  long long t;
  tmp++;
  vis[u] = tmp;
  cir.push(u);
  for (auto v : path[u]) {
    if (v == pa) continue;
    if (!vis[v])
      DFS(v, u);
    else if (vis[v] < vis[u]) {
      long long mx = u, mn = u;
      while (!cir.empty()) {
        t = cir.top();
        cir.pop();
        mx = max(mx, t);
        mn = min(mn, t);
        if (t == v) break;
      }
      p[mn] = mx - 1;
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    cin >> a >> b;
    path[a].push_back(b);
    path[b].push_back(a);
  }
  for (long long i = 1; i <= n; i++) {
    p[i] = n;
  }
  for (long long i = 1; i <= n; i++) {
    if (!vis[i]) DFS(i, 0);
  }
  tmp = n;
  for (long long i = n; i >= 1; i--) {
    tmp = min(tmp, p[i]);
    p[i] = tmp;
  }
  for (long long i = 1; i <= n; i++) {
    pre[i] = pre[i - 1] + p[i] - i + 1;
  }
  cin >> q;
  for (long long i = 0; i < q; i++) {
    cin >> a >> b;
    tmp = upper_bound(p + a, p + b + 1, b) - p;
    cout << (b - tmp + 2) * (b - tmp + 1) / 2 + pre[tmp - 1] - pre[a - 1]
         << endl;
  }
  return 0;
}

### Prompt

In CPP, your task is to solve the following problem:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long n, m, q;
long long a, b, tmp;
long long vis[300000 + 5];
vector<long long> path[300000 + 5];
stack<long long> cir;
long long p[300000 + 5], pre[300000 + 5];
void DFS(long long u, long long pa) {
  long long t;
  tmp++;
  vis[u] = tmp;
  cir.push(u);
  for (auto v : path[u]) {
    if (v == pa) continue;
    if (!vis[v])
      DFS(v, u);
    else if (vis[v] < vis[u]) {
      long long mx = u, mn = u;
      while (!cir.empty()) {
        t = cir.top();
        cir.pop();
        mx = max(mx, t);
        mn = min(mn, t);
        if (t == v) break;
      }
      p[mn] = mx - 1;
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    cin >> a >> b;
    path[a].push_back(b);
    path[b].push_back(a);
  }
  for (long long i = 1; i <= n; i++) {
    p[i] = n;
  }
  for (long long i = 1; i <= n; i++) {
    if (!vis[i]) DFS(i, 0);
  }
  tmp = n;
  for (long long i = n; i >= 1; i--) {
    tmp = min(tmp, p[i]);
    p[i] = tmp;
  }
  for (long long i = 1; i <= n; i++) {
    pre[i] = pre[i - 1] + p[i] - i + 1;
  }
  cin >> q;
  for (long long i = 0; i < q; i++) {
    cin >> a >> b;
    tmp = upper_bound(p + a, p + b + 1, b) - p;
    cout << (b - tmp + 2) * (b - tmp + 1) / 2 + pre[tmp - 1] - pre[a - 1]
         << endl;
  }
  return 0;
}

```