#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 500;
int n, m;
vector<int> G[maxn];
int visit[maxn] = {}, p[maxn] = {};
stack<int> st;
void dfs(int pos, int pre) {
  visit[pos] = 1;
  st.push(pos);
  for (int i : G[pos])
    if (i != pre) {
      if (visit[i] == 0) {
        dfs(i, pos);
      } else if (visit[i] == 1) {
        int mi = pos, ma = pos;
        while (!st.empty()) {
          int tp = st.top();
          st.pop();
          mi = min(mi, tp), ma = max(ma, tp);
          if (tp == i) break;
        }
        p[mi] = ma;
      }
    }
  visit[pos] = 2;
  if (!st.empty() && st.top() == pos) st.pop();
}
int bsearch(int l, int r, int val) {
  while (l < r) {
    int mid = (l + r + 1) >> 1;
    if (p[mid] > val) {
      r = mid - 1;
    } else
      l = mid;
  }
  return l;
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  for (int i = 1; i < n + 1; ++i) p[i] = n + 1;
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  for (int i = 1; i < n + 1; ++i)
    if (visit[i] != 2) {
      dfs(i, 0);
    }
  for (int i = n - 1; i >= 1; --i) {
    p[i] = min(p[i], p[i + 1]);
  }
  int q;
  cin >> q;
  while (q--) {
    unsigned long long a, b, lw, len, sum = 0;
    cin >> a >> b;
    while (1) {
      lw = bsearch(a, b, b);
      len = b - p[lw] + 1;
      if (p[lw] > b) {
        len = b - a + 1;
        sum += (len + 1) * len / 2;
        break;
      } else {
        sum += (len + 1) * len / 2 + len * (p[lw] - lw - 1);
        b = p[lw] - 1;
      }
    }
    cout << sum << '\n';
  }
}

### Prompt

Please create a solution in cpp 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 maxn = 3e5 + 500;
int n, m;
vector<int> G[maxn];
int visit[maxn] = {}, p[maxn] = {};
stack<int> st;
void dfs(int pos, int pre) {
  visit[pos] = 1;
  st.push(pos);
  for (int i : G[pos])
    if (i != pre) {
      if (visit[i] == 0) {
        dfs(i, pos);
      } else if (visit[i] == 1) {
        int mi = pos, ma = pos;
        while (!st.empty()) {
          int tp = st.top();
          st.pop();
          mi = min(mi, tp), ma = max(ma, tp);
          if (tp == i) break;
        }
        p[mi] = ma;
      }
    }
  visit[pos] = 2;
  if (!st.empty() && st.top() == pos) st.pop();
}
int bsearch(int l, int r, int val) {
  while (l < r) {
    int mid = (l + r + 1) >> 1;
    if (p[mid] > val) {
      r = mid - 1;
    } else
      l = mid;
  }
  return l;
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  for (int i = 1; i < n + 1; ++i) p[i] = n + 1;
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  for (int i = 1; i < n + 1; ++i)
    if (visit[i] != 2) {
      dfs(i, 0);
    }
  for (int i = n - 1; i >= 1; --i) {
    p[i] = min(p[i], p[i + 1]);
  }
  int q;
  cin >> q;
  while (q--) {
    unsigned long long a, b, lw, len, sum = 0;
    cin >> a >> b;
    while (1) {
      lw = bsearch(a, b, b);
      len = b - p[lw] + 1;
      if (p[lw] > b) {
        len = b - a + 1;
        sum += (len + 1) * len / 2;
        break;
      } else {
        sum += (len + 1) * len / 2 + len * (p[lw] - lw - 1);
        b = p[lw] - 1;
      }
    }
    cout << sum << '\n';
  }
}

```