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;
vector<int> max_len;
vector<int> visited;
vector<int> path;
void dfs(vector<vector<int> > &adj, int curr, int prev = -1) {
  visited[curr] = 1;
  path.push_back(curr);
  for (auto neigh : adj[curr]) {
    if (neigh == prev) continue;
    if (visited[neigh] == 1) {
      int mx_vert = 0;
      int min_vert = INT_MAX;
      for (int j = path.size() - 1; j >= 0; j--) {
        mx_vert = max(path[j], mx_vert);
        min_vert = min(path[j], min_vert);
        if (path[j] == neigh) break;
      }
      max_len[min_vert] = mx_vert - 1;
      continue;
    }
    if (visited[neigh] == 2) continue;
    dfs(adj, neigh, curr);
  }
  path.pop_back();
  visited[curr] = 2;
}
void solve() {
  int n, m;
  cin >> n >> m;
  max_len.resize(n + 1, n);
  max_len[0] = 0;
  visited.resize(n + 1, false);
  vector<vector<int> > adj(n + 1);
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  for (int i = 1; i <= n; i++)
    if (not visited[i]) dfs(adj, i);
  for (int i = n - 1; i >= 1; i--) max_len[i] = min(max_len[i + 1], max_len[i]);
  vector<long long> pref_sum(n + 1, 0);
  for (int i = 1; i <= n; i++) pref_sum[i] = pref_sum[i - 1] + max_len[i];
  vector<long long> int_sum(n + 1, 0);
  for (int i = 1; i <= n; i++) int_sum[i] = int_sum[i - 1] + i;
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    int break_idx = upper_bound(max_len.begin() + l, max_len.end(), r - 1) -
                    max_len.begin();
    long long sum1 = pref_sum[break_idx - 1] - pref_sum[l - 1] -
                     (int_sum[break_idx - 1] - int_sum[l - 1]) +
                     (break_idx - 1 - l + 1);
    long long sum2 = 1ll * r * (r - break_idx + 1) -
                     (int_sum[r] - int_sum[break_idx - 1]) +
                     (r - break_idx + 1);
    long long sum = sum1 + sum2;
    cout << sum << "\n";
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  solve();
  return 0;
}
