#include <bits/stdc++.h>
using namespace std;
const int nn = 3e5 + 8;
vector<int> G[nn];
int n, m;
int R[nn], pa[nn];
int trv[nn];
long long sum[nn], sum2[nn];
void dfs(int u) {
  trv[u] = 1;
  for (int v : G[u]) {
    if (v == pa[u]) continue;
    if (!trv[v]) {
      pa[v] = u;
      dfs(v);
    } else if (trv[v] == 1) {
      int Max = v, Min = v;
      for (int i = u; i != v; i = pa[i]) {
        Max = max(Max, i);
        Min = min(Min, i);
      }
      R[Min] = Max - 1;
    }
  }
  trv[u] = 2;
}
int _find(int x, int l, int r) {
  int res = l - 1;
  int y = r;
  while (l <= r) {
    int m = (l + r) >> 1;
    if (R[m] <= y) {
      res = m;
      l = m + 1;
    } else
      r = m - 1;
  }
  return res;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) R[i] = nn;
  for (int i = 1; i <= n; i++)
    if (!trv[i]) dfs(i);
  R[n] = n;
  for (int i = n - 1; i >= 1; i--) R[i] = min(R[i], R[i + 1]);
  for (int i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + (long long)(R[i] - i + 1);
    sum2[i] = sum2[i - 1] + i;
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    int x = _find(r, l, r);
    long long res =
        sum[x] - sum[l - 1] + (long long)(r + 1) * (r - x) - sum2[r] + sum2[x];
    printf("%lld\n", res);
  }
}

### Prompt

Your challenge is to write 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;
const int nn = 3e5 + 8;
vector<int> G[nn];
int n, m;
int R[nn], pa[nn];
int trv[nn];
long long sum[nn], sum2[nn];
void dfs(int u) {
  trv[u] = 1;
  for (int v : G[u]) {
    if (v == pa[u]) continue;
    if (!trv[v]) {
      pa[v] = u;
      dfs(v);
    } else if (trv[v] == 1) {
      int Max = v, Min = v;
      for (int i = u; i != v; i = pa[i]) {
        Max = max(Max, i);
        Min = min(Min, i);
      }
      R[Min] = Max - 1;
    }
  }
  trv[u] = 2;
}
int _find(int x, int l, int r) {
  int res = l - 1;
  int y = r;
  while (l <= r) {
    int m = (l + r) >> 1;
    if (R[m] <= y) {
      res = m;
      l = m + 1;
    } else
      r = m - 1;
  }
  return res;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) R[i] = nn;
  for (int i = 1; i <= n; i++)
    if (!trv[i]) dfs(i);
  R[n] = n;
  for (int i = n - 1; i >= 1; i--) R[i] = min(R[i], R[i + 1]);
  for (int i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + (long long)(R[i] - i + 1);
    sum2[i] = sum2[i - 1] + i;
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    int x = _find(r, l, r);
    long long res =
        sum[x] - sum[l - 1] + (long long)(r + 1) * (r - x) - sum2[r] + sum2[x];
    printf("%lld\n", res);
  }
}

```