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 long long N = 3e5 + 5;
long long read() {
  long long s = 0;
  char c = getchar(), lc = '+';
  while (c < '0' || '9' < c) lc = c, c = getchar();
  while ('0' <= c && c <= '9') s = s * 10 + c - '0', c = getchar();
  return lc == '-' ? -s : s;
}
void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x < 10)
    putchar(x + '0');
  else
    write(x / 10), putchar(x % 10 + '0');
}
void print(long long x = -1, char c = '\n') {
  write(x);
  putchar(c);
}
struct edge {
  long long to, nxt;
} e[N * 2];
long long head[N], cnte = 0;
void add_edge(long long u, long long v) {
  e[++cnte].to = v;
  e[cnte].nxt = head[u];
  head[u] = cnte;
}
long long ri[N], sum[N], fa[N], deep[N];
long long sigma(long long x) { return x * (x + 1) / 2; }
void dfs(long long now, long long father) {
  fa[now] = father;
  deep[now] = deep[father] + 1;
  for (long long i = head[now]; i; i = e[i].nxt) {
    long long to = e[i].to;
    if (to == father) continue;
    if (fa[to]) {
      if (deep[to] > deep[now]) continue;
      long long l = to, r = to;
      for (long long k = now; k != to; k = fa[k]) l = min(l, k), r = max(r, k);
      ri[l] = r - 1;
    } else
      dfs(to, now);
  }
}
signed main(signed Goodbye, char *Wangang[]) {
  (void)Goodbye, (void)Wangang;
  long long n = read(), m = read();
  for (long long i = 1; i <= m; i++) {
    long long u = read(), v = read();
    add_edge(u, v);
    add_edge(v, u);
  }
  for (long long i = 1; i <= n + 1; i++) ri[i] = n;
  for (long long i = 1; i <= n; i++)
    if (!fa[i]) dfs(i, -1);
  for (long long i = n; i >= 1; i--) ri[i] = min(ri[i], ri[i + 1]);
  for (long long i = n; i >= 1; i--) sum[i] = sum[i + 1] + ri[i] - i + 1;
  long long q = read();
  while (q--) {
    long long l = read(), r = read();
    long long pos = upper_bound(ri + l, ri + r + 1, r) - ri;
    print(sum[l] - sum[pos] + sigma(r - pos + 1));
  }
  return 0;
}
