#include <bits/stdc++.h>
using namespace std;
long long n, m, q, ans, rightEnd[300010], pref[300010], inTime[300010],
    outTime[300010], t;
stack<long long> st;
vector<vector<long long> > gr;
void dfs(long long x, long long par) {
  ++t;
  inTime[x] = t;
  st.push(x);
  for (auto i : gr[x]) {
    if (i == par) {
      continue;
    }
    if (!outTime[i] and inTime[i] and inTime[x] > inTime[i]) {
      long long tmpMin = i, tmpMax = i;
      while (!st.empty()) {
        long long tmp = st.top();
        tmpMin = min(tmpMin, tmp);
        tmpMax = max(tmpMax, tmp);
        st.pop();
        if (tmp == i) {
          break;
        }
      }
      rightEnd[tmpMin] = tmpMax;
    } else if (!inTime[i]) {
      dfs(i, x);
    }
  }
  ++t;
  outTime[x] = t;
  if (!st.empty() and st.top() == x) {
    st.pop();
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  cin >> n >> m;
  gr.resize(n + 1, vector<long long>());
  for (long long i = 0; i < m; ++i) {
    long long x, y;
    cin >> x >> y;
    gr[x].push_back(y);
    gr[y].push_back(x);
  }
  for (long long i = 1; i <= n; ++i) {
    rightEnd[i] = n + 2;
  }
  for (long long i = 1; i <= n; ++i) {
    if (!inTime[i]) {
      dfs(i, -1);
    }
  }
  for (long long i = n - 1; i; --i) {
    rightEnd[i] = min(rightEnd[i + 1], rightEnd[i]);
  }
  for (long long i = 1; i <= n; ++i) {
    pref[i] = pref[i - 1] + rightEnd[i] - i;
  }
  cin >> q;
  while (q--) {
    long long l, r;
    cin >> l >> r;
    long long l1 = l, r1 = r, tmp = r;
    while (l1 <= r1) {
      long long mid = (l1 + r1) / 2;
      if (rightEnd[mid] > r) {
        r1 = mid - 1;
        tmp = mid;
      } else {
        l1 = mid + 1;
      }
    }
    ans = pref[tmp - 1] - pref[l - 1];
    ans += ((r - tmp + 1) * (r - tmp + 2)) / 2;
    cout << ans << '\n';
  }
}

### Prompt

Generate 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;
long long n, m, q, ans, rightEnd[300010], pref[300010], inTime[300010],
    outTime[300010], t;
stack<long long> st;
vector<vector<long long> > gr;
void dfs(long long x, long long par) {
  ++t;
  inTime[x] = t;
  st.push(x);
  for (auto i : gr[x]) {
    if (i == par) {
      continue;
    }
    if (!outTime[i] and inTime[i] and inTime[x] > inTime[i]) {
      long long tmpMin = i, tmpMax = i;
      while (!st.empty()) {
        long long tmp = st.top();
        tmpMin = min(tmpMin, tmp);
        tmpMax = max(tmpMax, tmp);
        st.pop();
        if (tmp == i) {
          break;
        }
      }
      rightEnd[tmpMin] = tmpMax;
    } else if (!inTime[i]) {
      dfs(i, x);
    }
  }
  ++t;
  outTime[x] = t;
  if (!st.empty() and st.top() == x) {
    st.pop();
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  cin >> n >> m;
  gr.resize(n + 1, vector<long long>());
  for (long long i = 0; i < m; ++i) {
    long long x, y;
    cin >> x >> y;
    gr[x].push_back(y);
    gr[y].push_back(x);
  }
  for (long long i = 1; i <= n; ++i) {
    rightEnd[i] = n + 2;
  }
  for (long long i = 1; i <= n; ++i) {
    if (!inTime[i]) {
      dfs(i, -1);
    }
  }
  for (long long i = n - 1; i; --i) {
    rightEnd[i] = min(rightEnd[i + 1], rightEnd[i]);
  }
  for (long long i = 1; i <= n; ++i) {
    pref[i] = pref[i - 1] + rightEnd[i] - i;
  }
  cin >> q;
  while (q--) {
    long long l, r;
    cin >> l >> r;
    long long l1 = l, r1 = r, tmp = r;
    while (l1 <= r1) {
      long long mid = (l1 + r1) / 2;
      if (rightEnd[mid] > r) {
        r1 = mid - 1;
        tmp = mid;
      } else {
        l1 = mid + 1;
      }
    }
    ans = pref[tmp - 1] - pref[l - 1];
    ans += ((r - tmp + 1) * (r - tmp + 2)) / 2;
    cout << ans << '\n';
  }
}

```