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 INF = 1 << 30;
const long long inf = 1LL << 60;
const int N = 300010;
struct Edge {
  int to, next;
} e[N * 2];
int lst[N], d;
void add(int x, int y) { e[d].to = y, e[d].next = lst[x], lst[x] = d++; }
int read() {
  int w(0), f(0);
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') f = 1, c = getchar();
  while (c >= '0' && c <= '9') w = w * 10 + c - '0', c = getchar();
  return f ? -w : w;
}
int n, m, q, rm[N];
bool in[N], vis[N];
struct Pr {
  int l, r;
} s[N * 2];
int sz, fa[N], col[N];
void dfs(int t) {
  vis[t] = 1;
  for (int i = lst[t]; i >= 0; i = e[i].next)
    if (!vis[e[i].to]) {
      fa[e[i].to] = t;
      dfs(e[i].to);
    } else if (e[i].to != fa[t] && !col[e[i].to]) {
      ++sz;
      s[sz].l = s[sz].r = t;
      col[t] = sz;
      for (int j = t; j != e[i].to; j = fa[j]) {
        col[fa[j]] = sz;
        s[sz].l = min(s[sz].l, fa[j]);
        s[sz].r = max(s[sz].r, fa[j]);
      }
    }
}
bool cmp(const Pr& x, const Pr& y) { return x.r < y.r; }
long long sum[N];
int main() {
  n = read(), m = read();
  int x, y;
  memset(lst, -1, 4 * (n + 1)), d = 0;
  for (int i = 1; i <= m; ++i) {
    x = read(), y = read();
    add(x, y), add(y, x);
  }
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) dfs(i);
  sort(s + 1, s + sz + 1, cmp);
  int p = 1;
  long long tmp, ans;
  for (int i = 1; i <= n; ++i) {
    while (p <= sz && s[p].l < i) p++;
    if (p > sz)
      rm[i] = n;
    else
      rm[i] = s[p].r - 1;
    sum[i] = sum[i - 1] + (rm[i] - i + 1);
  }
  q = read();
  for (int i = 1, l, r, mid; i <= q; ++i) {
    x = read(), y = read();
    for (l = x, r = y, mid = (l + r) >> 1; l <= r; mid = (l + r) >> 1)
      if (rm[mid] < y)
        l = mid + 1;
      else
        r = mid - 1;
    tmp = y - l + 1;
    ans = sum[l - 1] - sum[x - 1] + tmp * (tmp + 1) / 2;
    printf("%I64d\n", ans);
  }
  return 0;
}
