#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
vector<int> adj[N];
int vis[N];
vector<int> points;
int goTill[N];
long long sum[N];
void dfs(int v, int p) {
  vis[v] = 1;
  points.push_back(v);
  for (auto it : adj[v]) {
    if (it == p || vis[it] == 2) continue;
    if (!vis[it]) {
      dfs(it, v);
      continue;
    }
    int mn = v, mx = v;
    for (auto pt = points.rbegin();; pt++) {
      mn = min(mn, *pt);
      mx = max(mx, *pt);
      if (*pt == it) break;
    }
    goTill[mn] = min(goTill[mn], mx - 1);
  }
  points.pop_back();
  vis[v] = 2;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  while (m--) {
    int ai, bi;
    cin >> ai >> bi;
    adj[ai].push_back(bi);
    adj[bi].push_back(ai);
  }
  for (int i = 1; i <= n; i++) {
    goTill[i] = n;
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) dfs(i, 0);
  }
  for (int i = n - 1; i >= 1; i--) {
    goTill[i] = min(goTill[i], goTill[i + 1]);
  }
  for (int i = 1; i <= n; i++) {
    sum[i] = goTill[i] - i + 1;
    sum[i] += sum[i - 1];
  }
  int q;
  cin >> q;
  while (q--) {
    int li, ri;
    cin >> li >> ri;
    int p = lower_bound(goTill + li, goTill + ri, ri) - goTill;
    long long ans = sum[p - 1] - sum[li - 1];
    long long cnt = ri - p + 1;
    cnt = (cnt * (cnt + 1)) / 2;
    ans += cnt;
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
const int N = 3e5 + 5;
vector<int> adj[N];
int vis[N];
vector<int> points;
int goTill[N];
long long sum[N];
void dfs(int v, int p) {
  vis[v] = 1;
  points.push_back(v);
  for (auto it : adj[v]) {
    if (it == p || vis[it] == 2) continue;
    if (!vis[it]) {
      dfs(it, v);
      continue;
    }
    int mn = v, mx = v;
    for (auto pt = points.rbegin();; pt++) {
      mn = min(mn, *pt);
      mx = max(mx, *pt);
      if (*pt == it) break;
    }
    goTill[mn] = min(goTill[mn], mx - 1);
  }
  points.pop_back();
  vis[v] = 2;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  while (m--) {
    int ai, bi;
    cin >> ai >> bi;
    adj[ai].push_back(bi);
    adj[bi].push_back(ai);
  }
  for (int i = 1; i <= n; i++) {
    goTill[i] = n;
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) dfs(i, 0);
  }
  for (int i = n - 1; i >= 1; i--) {
    goTill[i] = min(goTill[i], goTill[i + 1]);
  }
  for (int i = 1; i <= n; i++) {
    sum[i] = goTill[i] - i + 1;
    sum[i] += sum[i - 1];
  }
  int q;
  cin >> q;
  while (q--) {
    int li, ri;
    cin >> li >> ri;
    int p = lower_bound(goTill + li, goTill + ri, ri) - goTill;
    long long ans = sum[p - 1] - sum[li - 1];
    long long cnt = ri - p + 1;
    cnt = (cnt * (cnt + 1)) / 2;
    ans += cnt;
    cout << ans << "\n";
  }
  return 0;
}

```