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 N = 3e5 + 1;
int n, m, q, mark[N], bounds[N], lb[N];
vector<int> g[N];
long long ls[N], gr[N];
stack<int> st;
void dfs(int u, int par) {
  mark[u] = 1;
  st.push(u);
  for (int v : g[u])
    if (v != par) {
      if (!mark[v])
        dfs(v, u);
      else if (mark[v] == 1) {
        stack<int> tmp;
        int mn = v, mx = v;
        while (st.top() != v) {
          mn = min(st.top(), mn);
          mx = max(st.top(), mx);
          tmp.push(st.top());
          st.pop();
        }
        while (!tmp.empty()) st.push(tmp.top()), tmp.pop();
        bounds[mx] = min(mn, bounds[mx]);
      }
    }
  mark[u] = 2;
  st.pop();
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0, u, v; i < m; i++) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  fill(bounds + 1, bounds + n + 1, INT_MAX);
  for (int i = 1; i <= n; i++)
    if (!mark[i]) dfs(i, -1);
  for (int i = 1, j = 1; i <= n; i++) {
    j = max(bounds[i] + 1, j);
    lb[i] = j;
    ls[i] = ls[i - 1] + i;
    gr[i] = gr[i - 1] + i - j + 1;
  }
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    int s = l, e = r;
    while (s < e) {
      int mid = s + e + 1 >> 1;
      if (lb[mid] < l)
        s = mid;
      else
        e = mid - 1;
    }
    long long res = 0;
    if (lb[s] < l) {
      res += ls[s] - ls[l - 1] - 1ll * (s - l + 1) * (l - 1);
      res += gr[r] - gr[s];
    } else {
      res += gr[r] - gr[l - 1];
    }
    cout << res << '\n';
  }
}
