#include <bits/stdc++.h>
using namespace std;
int N, M, Q, vis[300009] = {};
long long int F[300009], last[300009];
vector<int> adjList[300009];
int vals[300009];
void dfs(int u, int p, int idx) {
  vis[u] = 1;
  vals[idx] = u;
  for (auto v : adjList[u]) {
    if (v == p || vis[v] == 2) continue;
    if (vis[v] == 1) {
      int wmax = u, wmin = u;
      for (int i = idx - 1; i >= 0; --i) {
        wmax = max(vals[i], wmax);
        wmin = min(vals[i], wmin);
        if (vals[i] == v) break;
      }
      last[wmax] = wmin;
    } else {
      dfs(v, u, idx + 1);
    }
  }
  vis[u] = 2;
}
int bs(int l, int r) {
  int mid;
  int a = l;
  while (l < r - 1) {
    mid = (l + r) >> 1;
    if (last[mid] < a)
      l = mid;
    else
      r = mid - 1;
  }
  if (last[r] < a) return r;
  return last[l] < a ? l : l - 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> N >> M;
  long long int a, b, dx;
  for (int i = 0; i < M; ++i) {
    cin >> a >> b;
    a--, b--;
    adjList[a].push_back(b);
    adjList[b].push_back(a);
  }
  cin >> Q;
  memset(last, -1, sizeof last);
  for (int i = 0; i < N; ++i) {
    if (!vis[i]) dfs(i, -1, 0);
  }
  F[0] = 1;
  for (int i = 1; i < N; ++i) {
    last[i] = max(last[i], last[i - 1]);
    F[i] = F[i - 1] + i - last[i];
  }
  while (Q--) {
    cin >> a >> b;
    --a, --b;
    int l = a;
    a = bs(a, b);
    dx = a - l + 1;
    dx = dx * (dx + 1) >> 1;
    cout << dx + F[b] - F[a] << '\n';
  }
}

### Prompt

Create a solution in cpp for 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;
int N, M, Q, vis[300009] = {};
long long int F[300009], last[300009];
vector<int> adjList[300009];
int vals[300009];
void dfs(int u, int p, int idx) {
  vis[u] = 1;
  vals[idx] = u;
  for (auto v : adjList[u]) {
    if (v == p || vis[v] == 2) continue;
    if (vis[v] == 1) {
      int wmax = u, wmin = u;
      for (int i = idx - 1; i >= 0; --i) {
        wmax = max(vals[i], wmax);
        wmin = min(vals[i], wmin);
        if (vals[i] == v) break;
      }
      last[wmax] = wmin;
    } else {
      dfs(v, u, idx + 1);
    }
  }
  vis[u] = 2;
}
int bs(int l, int r) {
  int mid;
  int a = l;
  while (l < r - 1) {
    mid = (l + r) >> 1;
    if (last[mid] < a)
      l = mid;
    else
      r = mid - 1;
  }
  if (last[r] < a) return r;
  return last[l] < a ? l : l - 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> N >> M;
  long long int a, b, dx;
  for (int i = 0; i < M; ++i) {
    cin >> a >> b;
    a--, b--;
    adjList[a].push_back(b);
    adjList[b].push_back(a);
  }
  cin >> Q;
  memset(last, -1, sizeof last);
  for (int i = 0; i < N; ++i) {
    if (!vis[i]) dfs(i, -1, 0);
  }
  F[0] = 1;
  for (int i = 1; i < N; ++i) {
    last[i] = max(last[i], last[i - 1]);
    F[i] = F[i - 1] + i - last[i];
  }
  while (Q--) {
    cin >> a >> b;
    --a, --b;
    int l = a;
    a = bs(a, b);
    dx = a - l + 1;
    dx = dx * (dx + 1) >> 1;
    cout << dx + F[b] - F[a] << '\n';
  }
}

```