#include <bits/stdc++.h>
using namespace std;
long long n, m, a[300005], dp[300005], sm[300005];
vector<long long> adj[300005];
long long vis[300005];
stack<long long> s;
void dfs(long long i, long long p) {
  vis[i] = 1;
  s.push(i);
  long long mx = LLONG_MIN, mn = LLONG_MAX;
  for (auto j : adj[i]) {
    if (!vis[j])
      dfs(j, i);
    else if (vis[j] == 1 && j != p) {
      while (s.top() != j) {
        mx = max(mx, s.top());
        mn = min(mn, s.top());
        s.pop();
      }
      mx = max(mx, s.top());
      mn = min(mn, s.top());
      s.pop();
      a[mn] = min(a[mn], mx);
      vis[i] = 2;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  memset(a, 1000000007, sizeof(a));
  long long u, v;
  cin >> n >> m;
  for (long long i = (long long)0; i < (long long)m; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (long long i = (long long)1; i < (long long)n + 1; i++) {
    if (!vis[i]) dfs(i, -1);
  }
  dp[n] = n;
  long long mn = n;
  for (long long i = (long long)n - 1; i >= (long long)1; i--) {
    mn = min(mn, a[i] - 1);
    dp[i] = mn;
  }
  sm[1] = dp[1];
  for (long long i = (long long)2; i < (long long)n + 1; i++)
    sm[i] = sm[i - 1] + dp[i];
  long long q, l, r, ans;
  cin >> q;
  while (q--) {
    cin >> l >> r;
    ans = 0;
    long long p = upper_bound(dp + l, dp + r + 1, r) - dp;
    ans = (r + 1) * (r - p + 1) - r * (r + 1) / 2 + p * (p - 1) / 2;
    ans += (sm[p - 1] - sm[l - 1] + p - l - p * (p - 1) / 2 + l * (l - 1) / 2);
    cout << ans << "\n";
  }
}

### Prompt

Please formulate a cpp solution to 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, a[300005], dp[300005], sm[300005];
vector<long long> adj[300005];
long long vis[300005];
stack<long long> s;
void dfs(long long i, long long p) {
  vis[i] = 1;
  s.push(i);
  long long mx = LLONG_MIN, mn = LLONG_MAX;
  for (auto j : adj[i]) {
    if (!vis[j])
      dfs(j, i);
    else if (vis[j] == 1 && j != p) {
      while (s.top() != j) {
        mx = max(mx, s.top());
        mn = min(mn, s.top());
        s.pop();
      }
      mx = max(mx, s.top());
      mn = min(mn, s.top());
      s.pop();
      a[mn] = min(a[mn], mx);
      vis[i] = 2;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  memset(a, 1000000007, sizeof(a));
  long long u, v;
  cin >> n >> m;
  for (long long i = (long long)0; i < (long long)m; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (long long i = (long long)1; i < (long long)n + 1; i++) {
    if (!vis[i]) dfs(i, -1);
  }
  dp[n] = n;
  long long mn = n;
  for (long long i = (long long)n - 1; i >= (long long)1; i--) {
    mn = min(mn, a[i] - 1);
    dp[i] = mn;
  }
  sm[1] = dp[1];
  for (long long i = (long long)2; i < (long long)n + 1; i++)
    sm[i] = sm[i - 1] + dp[i];
  long long q, l, r, ans;
  cin >> q;
  while (q--) {
    cin >> l >> r;
    ans = 0;
    long long p = upper_bound(dp + l, dp + r + 1, r) - dp;
    ans = (r + 1) * (r - p + 1) - r * (r + 1) / 2 + p * (p - 1) / 2;
    ans += (sm[p - 1] - sm[l - 1] + p - l - p * (p - 1) / 2 + l * (l - 1) / 2);
    cout << ans << "\n";
  }
}

```