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;
long long n, m, q, i, j, a, b, l, r, t, cnt, L, R, mid, sum[400000];
long long used[400000], p[400000], mx[400000];
vector<int> g[400000];
int cycle[400000], sz[400000];
void dfs(int v) {
  int i, to;
  used[v] = 1;
  for (i = 0; i < g[v].size(); i++) {
    to = g[v][i];
    if (p[v] == to) continue;
    if (used[to] == 1 && cycle[v] == -1) {
      int u = v, cnt = 1;
      while (u != to) {
        cycle[u] = t;
        cnt++;
        u = p[u];
      }
      cycle[to] = t;
      sz[t] = cnt;
      t++;
    }
    if (!used[to]) {
      p[to] = v;
      dfs(to);
    }
  }
  used[v] = 2;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (i = 0; i < m; i++) {
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  fill(cycle, cycle + n, -1);
  for (i = 0; i < n; i++) {
    if (!used[i]) dfs(i);
  }
  j = 0;
  for (i = 0; i < n; i++) {
    while (j < n && (cycle[j] == -1 || sz[cycle[j]] != 1)) {
      if (cycle[j] != -1) sz[cycle[j]]--;
      j++;
    }
    mx[i] = j - 1;
    sum[i] = ((i) ? sum[i - 1] : 0) + (mx[i] - i + 1);
    if (cycle[i] != -1) sz[cycle[i]]++;
  }
  cin >> q;
  for (i = 0; i < q; i++) {
    cin >> l >> r;
    l--;
    r--;
    L = l;
    R = r;
    while (L < R) {
      mid = (L + R) / 2;
      if (mx[mid] <= r)
        L = mid + 1;
      else
        R = mid;
    }
    cout << (r - L + 1) * (r - L + 2) / 2 + ((L) ? sum[L - 1] : 0) -
                ((l) ? sum[l - 1] : 0)
         << endl;
  }
  return 0;
}
