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>
#include <bits/stdc++.h>
using namespace std;
const int INF = 500000;
vector<vector<int> > g;
vector<bool> use;
vector<int> path;
vector<pair<int, int> > cycles;
void dfs(int v, int parent) {
  use[v] = true;
  path.push_back(v);
  for (int i = 0; i < g[v].size(); i++) {
    int u = g[v][i];
    if (u != parent && use[u]) {
      int start = v;
      int end = v;
      for (int j = path.size() - 1; j >= 0; j--) {
        start = min(start, path[j]);
        end = max(end, path[j]);
        if (path[j] == u) break;
      }
      cycles.push_back(make_pair(start, end));
    }
  }
  for (int i = 0; i < g[v].size(); i++) {
    int u = g[v][i];
    if (!use[u]) {
      dfs(u, v);
    }
  }
  path.pop_back();
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  g.resize(n + 1);
  for (int i = 0; i < m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  int q;
  scanf("%d", &q);
  vector<pair<int, int> > queries(q);
  for (int i = 0; i < q; i++)
    scanf("%d%d", &queries[i].first, &queries[i].second);
  use.resize(n + 1, false);
  path.reserve(n + 1);
  for (int i = 1; i <= n; i++) {
    if (!use[i]) {
      dfs(i, -1);
    }
  }
  cycles.push_back(make_pair(INF, INF));
  sort(cycles.begin(), cycles.end());
  int nCycles = cycles.size();
  vector<int> runningMin(nCycles);
  runningMin[nCycles - 1] = cycles[nCycles - 1].second;
  for (int i = nCycles - 2; i >= 0; i--) {
    runningMin[i] = min(runningMin[i + 1], cycles[i].second);
  }
  vector<int> mx(n + 1);
  vector<long long> prefMx(n + 1);
  prefMx[0] = 0;
  int cur = 0;
  for (int i = 1; i <= n; i++) {
    while (cycles[cur].first < i) cur++;
    mx[i] = runningMin[cur];
    prefMx[i] = prefMx[i - 1] + mx[i];
  }
  for (int z = 0; z < queries.size(); z++) {
    long long L = queries[z].first, R = queries[z].second;
    int ll = L, rr = R;
    while (ll <= rr) {
      int x = (ll + rr) / 2;
      if (mx[x] <= R)
        ll = x + 1;
      else
        rr = x - 1;
    }
    long long u = ll;
    long long result = (R - u + 1) * (R + 1) - (L + R) * (R - L + 1) / 2 +
                       prefMx[u - 1] - prefMx[L - 1];
    cout << result << endl;
  }
  return 0;
}
