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 MAXN = 3e5 + 13;
pair<int, pair<int, int> > pr;
bool odw[MAXN];
long long dp[MAXN];
int pocz[MAXN];
int oj[MAXN];
int step[MAXN];
int lft[MAXN];
vector<int> v[MAXN];
vector<int> tree[MAXN];
void DFS(int x) {
  step[x] = step[oj[x]] + 1;
  odw[x] = true;
  for (auto it : v[x]) {
    if (!odw[it]) {
      oj[it] = x;
      tree[x].push_back(it);
      DFS(it);
    }
  }
}
pair<int, pair<int, int> > DFS2(int x) {
  odw[x] = false;
  pair<int, pair<int, int> > pr2 = {0, {0, 0}};
  for (auto it : tree[x]) {
    if (odw[it]) {
      pr = DFS2(it);
      if (x == pr.first) {
        pr.second.first = min(pr.second.first, x);
        pr.second.second = max(pr.second.second, x);
        if (pocz[pr.second.first] == 0)
          pocz[pr.second.first] = pr.second.second;
        else
          pocz[pr.second.first] = min(pocz[pr.second.first], pr.second.second);
      } else if (pr.first != 0) {
        pr2 = pr;
        pr2.second.first = min(pr2.second.first, x);
        pr2.second.second = max(pr2.second.second, x);
      }
    } else
      pr2 = {it, {x, x}};
  }
  return pr2;
}
int main() {
  int n, m, z;
  int x, y;
  scanf("%d %d", &n, &m);
  while (m--) {
    scanf("%d %d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) {
    if (!odw[i]) DFS(i);
  }
  for (int i = 1; i <= n; i++) {
    for (auto it : v[i]) {
      if (step[i] > step[it] && oj[i] != it) tree[i].push_back(it);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (odw[i]) DFS2(i);
  }
  int mn = n + 1;
  for (int i = n; i > 0; i--) {
    if (pocz[i] != 0) {
      mn = min(mn, pocz[i]);
      lft[pocz[i]] = max(lft[pocz[i]], i);
    }
    dp[i] = (long long)(mn - i);
  }
  for (int i = 2; i <= n; i++) dp[i] += dp[i - 1];
  for (int i = 1; i <= n; i++) lft[i] = max(lft[i], lft[i - 1]);
  long long wyn, d;
  int p;
  scanf("%d", &z);
  while (z--) {
    scanf("%d %d", &x, &y);
    p = lft[y];
    d = y - max(x, p + 1) + 1;
    wyn = d * (d + 1) / 2;
    if (p >= x) wyn += dp[p] - dp[x - 1];
    printf("%lld\n", wyn);
  }
  return 0;
}
