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;
int n, m, color[1000000], used[1000000], a, b, mx[1000000], q;
long long dp[1000000];
vector<int> ord;
vector<pair<int, int>> cycles;
set<int> ss;
vector<int> g[1000000];
pair<int, int> findcycle(int l, int r, int y) {
  int qq = ord.size() - 1;
  while (y != ord[qq]) {
    l = min(ord[qq], l);
    r = max(ord[qq], r);
    qq--;
  }
  l = min(ord[qq], l);
  r = max(ord[qq], r);
  return make_pair(l, r);
}
void dfs(int x, int pr) {
  ord.push_back(x);
  used[x] = 1;
  for (int t = 0; t < g[x].size(); t++) {
    if (used[g[x][t]] == 1 && g[x][t] != pr) {
      cycles.push_back(findcycle(x, x, g[x][t]));
    }
    if (used[g[x][t]] == 0) dfs(g[x][t], x);
  }
  ord.pop_back();
  used[x] = 2;
}
long long solve(int x, int y) {
  while (x < y - 1) {
    int m = (x + y) / 2;
    if (mx[m] > b)
      y = m;
    else
      x = m;
  }
  long long otv, otv1;
  if (mx[x] > b)
    otv = 1LL * (b - x + 1) * (b - x + 2) / 2;
  else
    otv = dp[x] - dp[a - 1] + 1LL * (b - x) * (b - x + 1) / 2;
  return otv;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (int i = 1; i <= n; i++) {
    if (!used[i]) dfs(i, i);
  }
  for (int i = 0; i < cycles.size(); i++) {
    mx[cycles[i].first] = cycles[i].second;
  }
  ss.insert(-(n + 1));
  for (int i = n; i >= 1; i--) {
    if (mx[i]) ss.insert(-mx[i]);
    set<int>::iterator it = ss.end();
    it--;
    mx[i] = -(*it);
  }
  for (int i = 1; i <= n; i++) {
    dp[i] = dp[i - 1] + mx[i] - i;
  }
  cin >> q;
  for (int i = 0; i < q; i++) {
    cin >> a >> b;
    cout << solve(a, b) << endl;
  }
}
