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 + 10;
int n, m;
int nivel[maxn], pai[maxn];
bool mark[maxn];
int mx[maxn];
long long pref[maxn];
vector<int> grafo[maxn];
void get_ciclo(int u, int v) {
  int menor = n + 1, maior = 0;
  while (u && nivel[u] >= nivel[v]) {
    menor = min(menor, u);
    maior = max(maior, u);
    u = pai[u];
  }
  mx[menor] = min(mx[menor], maior);
}
void dfs(int u, int p) {
  mark[u] = 1;
  for (auto v : grafo[u]) {
    if (mark[v]) {
      if (v == p || nivel[v] > nivel[u]) continue;
      get_ciclo(u, v);
      continue;
    }
    nivel[v] = nivel[u] + 1;
    pai[v] = u;
    dfs(v, u);
  }
}
long long soma(int x) { return 1ll * x * (x + 1) / 2ll; }
int main(void) {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    grafo[u].push_back(v);
    grafo[v].push_back(u);
  }
  for (int i = 1; i <= n + 1; i++) mx[i] = n + 1;
  for (int i = 1; i <= n; i++)
    if (!mark[i]) dfs(i, 0);
  for (int i = n; i >= 1; i--) mx[i] = min(mx[i], mx[i + 1]);
  for (int i = 1; i <= n; i++) pref[i] = pref[i - 1] + 1ll * (mx[i] - i);
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    int ini = l, fim = r, ind = r + 1;
    while (ini <= fim) {
      int mid = (ini + fim) >> 1;
      if (mx[mid] > r)
        ind = mid, fim = mid - 1;
      else
        ini = mid + 1;
    }
    long long ans = 0;
    if (ind != r + 1) {
      ans += 1ll * (r + 1) * (r - ind + 1);
      ans -= (1ll * soma(r) - soma(ind - 1));
    }
    if (ind > l) ans += pref[ind - 1] - pref[l - 1];
    printf("%lld\n", ans);
  }
}
