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 = 310000;
vector<int> g[N];
struct Cyc {
  int mi, ma;
  Cyc() {
    this->mi = 0;
    this->ma = 0;
  }
  Cyc(int mi, int ma) {
    this->mi = mi;
    this->ma = ma;
  }
};
bool visited[N];
vector<int> pop;
bool inside[N];
int tercyc = 0;
int cycles[N];
int n, m;
void dfs(int a, int father) {
  visited[a] = true;
  inside[a] = true;
  pop.push_back(a);
  for (int i = 0; i < g[a].size(); i++) {
    int d = g[a][i];
    if (!visited[d]) {
      dfs(d, a);
    } else if (inside[d] and d != father) {
      int mn = d;
      int mi = d;
      for (int j = pop.size() - 1; pop[j] != d; j--) {
        mi = min(mi, pop[j]);
        mn = max(mn, pop[j]);
      }
      if (cycles[mn] != 0)
        cycles[mn] = min(mi, cycles[mn]);
      else
        cycles[mn] = mi;
    }
  }
  inside[a] = false;
  pop.pop_back();
}
int dp[N];
long long pref[N];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (int i = 1; i <= n; i++) {
    if (!visited[i]) dfs(i, -1);
  }
  int r = 1;
  for (int i = 1; i <= n; i++) {
    while ((r <= n - 1) and (cycles[r + 1] < i)) {
      r++;
    }
    dp[i] = r;
    pref[i] = pref[i - 1] + dp[i] - i + 1;
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int a, b;
    scanf("%d%d", &a, &b);
    int r = lower_bound(dp + a, dp + b + 1, b) - dp;
    long long wszystko = b - r + 1;
    printf("%lld\n",
           (wszystko * (wszystko + 1)) / 2 + pref[r - 1] - pref[a - 1]);
  }
}
