#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 7;
vector<int> G[N];
int vis[N] = {0}, p[N];
long long sum[N] = {0};
stack<int> s;
void dfs(int u, int fa) {
  s.push(u);
  vis[u] = 1;
  for (int i = 0; i < (int)G[u].size(); i++) {
    int v = G[u][i];
    if (v == fa) continue;
    if (!vis[v])
      dfs(v, u);
    else if (vis[v] == 1) {
      int maxx = u, minn = u;
      while (!s.empty()) {
        int t = s.top();
        s.pop();
        maxx = max(t, maxx);
        minn = min(t, minn);
        if (t == v) break;
      }
      p[minn] = maxx;
    }
  }
  if (!s.empty() && s.top() == u) s.pop();
  vis[u] = 2;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    G[x].push_back(y);
    G[y].push_back(x);
  }
  for (int i = 1; i <= n + 1; i++) p[i] = n + 1;
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i, -1);
  for (int i = n; i >= 1; i--) {
    p[i] = min(p[i], p[i + 1]);
    sum[i] = sum[i + 1] + p[i] - i;
  }
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    int L = l, R = r + 1;
    while (L < R) {
      int m = (L + R) / 2;
      if (p[m] > r)
        R = m;
      else
        L = m + 1;
    }
    cout << sum[l] - sum[L] + 1ll * (r - L + 1) * (r - L + 2) / 2 << endl;
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution 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 + 7;
vector<int> G[N];
int vis[N] = {0}, p[N];
long long sum[N] = {0};
stack<int> s;
void dfs(int u, int fa) {
  s.push(u);
  vis[u] = 1;
  for (int i = 0; i < (int)G[u].size(); i++) {
    int v = G[u][i];
    if (v == fa) continue;
    if (!vis[v])
      dfs(v, u);
    else if (vis[v] == 1) {
      int maxx = u, minn = u;
      while (!s.empty()) {
        int t = s.top();
        s.pop();
        maxx = max(t, maxx);
        minn = min(t, minn);
        if (t == v) break;
      }
      p[minn] = maxx;
    }
  }
  if (!s.empty() && s.top() == u) s.pop();
  vis[u] = 2;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    G[x].push_back(y);
    G[y].push_back(x);
  }
  for (int i = 1; i <= n + 1; i++) p[i] = n + 1;
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i, -1);
  for (int i = n; i >= 1; i--) {
    p[i] = min(p[i], p[i + 1]);
    sum[i] = sum[i + 1] + p[i] - i;
  }
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    int L = l, R = r + 1;
    while (L < R) {
      int m = (L + R) / 2;
      if (p[m] > r)
        R = m;
      else
        L = m + 1;
    }
    cout << sum[l] - sum[L] + 1ll * (r - L + 1) * (r - L + 2) / 2 << endl;
  }
  return 0;
}

```