#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:536870912")
#pragma warning(disable : 4996)
using namespace std;
const int N = 3e5 + 100;
vector<int> v[N];
int sz[N], col[N], cs[N], rb[N];
long long int ans[N], sum[N];
bool used[N];
int z = 1;
pair<int, int> dfs(int node, int anc) {
  used[node] = 1;
  pair<int, int> ret = {-1, -1};
  for (int to : v[node]) {
    if (to == anc) continue;
    if (used[to]) {
      if (ret.first == -1) ret = {to, z++};
    } else {
      auto g = dfs(to, node);
      if (g.first != -1) ret = g;
    }
  }
  if (ret.first != -1) {
    sz[ret.second]++;
    col[node] = ret.second;
    return ret.first == node ? make_pair(-1, -1) : ret;
  }
  return {-1, -1};
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--, b--;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  for (int i = 0; i < n; i++) {
    if (!used[i]) {
      dfs(i, -1);
    }
  }
  int r = -1, t = 0;
  for (int i = 0; i < n; i++) {
    while (!t && r < n) {
      r++;
      if (col[r]) {
        cs[col[r]]++;
        if (cs[col[r]] == sz[col[r]]) t = 1;
      }
    }
    rb[i] = r - 1;
    if (col[i]) {
      if (cs[col[i]] == sz[col[i]]) t = 0;
      cs[col[i]]--;
    }
  }
  for (int i = 0; i < n; i++) {
    sum[i + 1] = sum[i] + rb[i] - i + 1;
  }
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    x--, y--;
    long long int w = upper_bound(rb, rb + n, y) - rb - 1,
                  h = y - max(x - 1ll, w),
                  ans = sum[max((long long int)x, w + 1)] - sum[x] +
                        h * (h + 1) / 2;
    printf("%lld\n", ans);
  }
  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>
#pragma comment(linker, "/STACK:536870912")
#pragma warning(disable : 4996)
using namespace std;
const int N = 3e5 + 100;
vector<int> v[N];
int sz[N], col[N], cs[N], rb[N];
long long int ans[N], sum[N];
bool used[N];
int z = 1;
pair<int, int> dfs(int node, int anc) {
  used[node] = 1;
  pair<int, int> ret = {-1, -1};
  for (int to : v[node]) {
    if (to == anc) continue;
    if (used[to]) {
      if (ret.first == -1) ret = {to, z++};
    } else {
      auto g = dfs(to, node);
      if (g.first != -1) ret = g;
    }
  }
  if (ret.first != -1) {
    sz[ret.second]++;
    col[node] = ret.second;
    return ret.first == node ? make_pair(-1, -1) : ret;
  }
  return {-1, -1};
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--, b--;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  for (int i = 0; i < n; i++) {
    if (!used[i]) {
      dfs(i, -1);
    }
  }
  int r = -1, t = 0;
  for (int i = 0; i < n; i++) {
    while (!t && r < n) {
      r++;
      if (col[r]) {
        cs[col[r]]++;
        if (cs[col[r]] == sz[col[r]]) t = 1;
      }
    }
    rb[i] = r - 1;
    if (col[i]) {
      if (cs[col[i]] == sz[col[i]]) t = 0;
      cs[col[i]]--;
    }
  }
  for (int i = 0; i < n; i++) {
    sum[i + 1] = sum[i] + rb[i] - i + 1;
  }
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    x--, y--;
    long long int w = upper_bound(rb, rb + n, y) - rb - 1,
                  h = y - max(x - 1ll, w),
                  ans = sum[max((long long int)x, w + 1)] - sum[x] +
                        h * (h + 1) / 2;
    printf("%lld\n", ans);
  }
  return 0;
}

```