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 = (int)3e5 + 5;
vector<int> t[N], cykl[N];
int q, n, m, czas, low[N], dol[N], pre[N], ile[N], tym[N], tab[N], cykle;
long long s[N];
bool o[N];
void dfs(int v, int p) {
  czas++;
  pre[v] = czas;
  o[v] = true;
  for (int w : t[v])
    if (!o[w]) {
      dfs(w, v);
      if (pre[low[w]] <= pre[v]) {
        cykl[v].push_back(dol[w]);
        if (pre[low[w]] < pre[low[v]]) {
          low[v] = low[w];
          dol[v] = dol[w];
        }
      }
    } else if (pre[w] < pre[v] and w != p) {
      low[v] = w;
      dol[v] = v;
      cykl[v].push_back(v);
    }
}
void dodaj(int v) {
  for (int w : cykl[v]) {
    tym[w]++;
    if (tym[w] == ile[w]) cykle++;
  }
}
void usun(int v) {
  for (int w : cykl[v]) {
    if (tym[w] == ile[w]) cykle--;
    tym[w]--;
  }
}
int szukaj(int poc, int kon, int x) {
  int wyn = poc;
  while (poc <= kon) {
    int sr = (poc + kon) / 2;
    if (tab[sr] < x) {
      wyn = sr;
      poc = sr + 1;
    } else
      kon = sr - 1;
  }
  return wyn;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    t[a].push_back(b);
    t[b].push_back(a);
  }
  pre[0] = n + 1;
  for (int i = 1; i <= n; i++)
    if (!o[i]) dfs(i, 0);
  for (int i = 1; i <= n; i++)
    for (int w : cykl[i]) ile[w]++;
  int kon = 0;
  for (int poc = 1; poc <= n; poc++) {
    while (cykle == 0 and kon <= n) {
      kon++;
      dodaj(kon);
    }
    tab[poc] = kon - 1;
    usun(poc);
    s[poc] = s[poc - 1] + tab[poc];
  }
  cin >> q;
  while (q) {
    int a, b;
    cin >> a >> b;
    int poz = szukaj(a, b, b);
    long long l1 = b - a + 1, dl = poz - a + 1;
    long long rez = (l1 * (l1 + 1)) / 2;
    if (tab[poz] < b) rez += -(dl * (long long)b) + s[poz] - s[a - 1];
    cout << rez << "\n";
    q--;
  }
}
