#include <bits/stdc++.h>
using namespace std;
const int MAX = 300001;
int n, m;
vector<int> rightt(MAX, MAX + 10);
vector<vector<int> > adjList(MAX);
int UNVISITED = 0, EXPLORED = 1, VISITED = 2;
int visited[MAX];
stack<int> s;
void dfs(int node, int p) {
  s.push(node);
  visited[node] = EXPLORED;
  for (auto nxt : adjList[node])
    if (nxt != p) {
      if (!visited[nxt]) {
        dfs(nxt, node);
      } else if (visited[nxt] == EXPLORED) {
        int mn = MAX + 10, mx = -1;
        while (1) {
          int num = s.top();
          s.pop();
          mn = min(mn, num);
          mx = max(mx, num);
          if (num == nxt) break;
        }
        rightt[mn] = mx;
      }
    }
  visited[node] = VISITED;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  cin >> n >> m;
  while (m--) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    adjList[a].push_back(b);
    adjList[b].push_back(a);
  }
  for (int i = 0; i < n; i++)
    if (visited[i] == UNVISITED) dfs(i, -1);
  for (int i = n - 2; i >= 0; i--) rightt[i] = min(rightt[i], rightt[i + 1]);
  long long sumDiff[MAX + 1];
  sumDiff[0] = rightt[0];
  for (int i = 1; i < MAX; i++) sumDiff[i] = sumDiff[i - 1] + rightt[i] - i;
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    auto ptr = upper_bound(rightt.begin() + l, rightt.begin() + r, r);
    if (ptr == rightt.end()) {
      cout << sumDiff[r] - (l == 0 ? 0 : sumDiff[l - 1]) << "\n";
    } else {
      int start = ptr - rightt.begin();
      cout << 1ll * (r - start + 1) * (r - start + 2) / 2 +
                  (start == 0 ? 0 : sumDiff[start - 1]) -
                  (l == 0 ? 0 : sumDiff[l - 1])
           << "\n";
    }
  }
}

### 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 MAX = 300001;
int n, m;
vector<int> rightt(MAX, MAX + 10);
vector<vector<int> > adjList(MAX);
int UNVISITED = 0, EXPLORED = 1, VISITED = 2;
int visited[MAX];
stack<int> s;
void dfs(int node, int p) {
  s.push(node);
  visited[node] = EXPLORED;
  for (auto nxt : adjList[node])
    if (nxt != p) {
      if (!visited[nxt]) {
        dfs(nxt, node);
      } else if (visited[nxt] == EXPLORED) {
        int mn = MAX + 10, mx = -1;
        while (1) {
          int num = s.top();
          s.pop();
          mn = min(mn, num);
          mx = max(mx, num);
          if (num == nxt) break;
        }
        rightt[mn] = mx;
      }
    }
  visited[node] = VISITED;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  cin >> n >> m;
  while (m--) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    adjList[a].push_back(b);
    adjList[b].push_back(a);
  }
  for (int i = 0; i < n; i++)
    if (visited[i] == UNVISITED) dfs(i, -1);
  for (int i = n - 2; i >= 0; i--) rightt[i] = min(rightt[i], rightt[i + 1]);
  long long sumDiff[MAX + 1];
  sumDiff[0] = rightt[0];
  for (int i = 1; i < MAX; i++) sumDiff[i] = sumDiff[i - 1] + rightt[i] - i;
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    auto ptr = upper_bound(rightt.begin() + l, rightt.begin() + r, r);
    if (ptr == rightt.end()) {
      cout << sumDiff[r] - (l == 0 ? 0 : sumDiff[l - 1]) << "\n";
    } else {
      int start = ptr - rightt.begin();
      cout << 1ll * (r - start + 1) * (r - start + 2) / 2 +
                  (start == 0 ? 0 : sumDiff[start - 1]) -
                  (l == 0 ? 0 : sumDiff[l - 1])
           << "\n";
    }
  }
}

```