We have N colored balls arranged in a row from left to right; the color of the i-th ball from the left is c_i.

You are given Q queries. The i-th query is as follows: how many different colors do the l_i-th through r_i-th balls from the left have?

Constraints

* 1\leq N,Q \leq 5 \times 10^5
* 1\leq c_i \leq N
* 1\leq l_i \leq r_i \leq N
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N Q
c_1 c_2 \cdots c_N
l_1 r_1
l_2 r_2
:
l_Q r_Q


Output

Print Q lines. The i-th line should contain the response to the i-th query.

Examples

Input

4 3
1 2 1 3
1 3
2 4
3 3


Output

2
3
1


Input

10 10
2 5 6 5 2 1 7 9 7 2
5 5
2 4
6 7
2 2
7 8
7 9
1 8
6 9
8 10
6 8


Output

1
2
2
1
2
2
6
3
3
3
#include <iostream>
#include <cstdio>
#include <algorithm>
#define N_MAX 5000000
using namespace std;

struct node
{
  int l, r, id;
} q[N_MAX + 5];

// c:binary indexed tree, f:last ingdex(j) of a[j]
int N, a[N_MAX + 5], m, c[N_MAX + 5], f[N_MAX + 5], ans[N_MAX + 5];

int lowbit(int x) {return x & (-x);}
int cmp(node x, node y) {return x.r < y.r;}
void add(int x, int k){
  for (; x <= N; x += lowbit(x))
    c[x] += k;
}
int sum(int x)
{
  int ans = 0;
  for (; x; x -= lowbit(x))
    ans += c[x];
  return ans;
}

int main()
{
  scanf("%d%d", &N, &m);
  for (int i = 1; i <= N; i++)
    scanf("%d", &a[i]);
  for (int i = 1; i <= m; i++)
    scanf("%d%d", &q[i].l, &q[i].r), q[i].id = i;
  sort(q + 1, q + m + 1, cmp);
  for (int i = 1; i <= m; i++)
  {
    for (int j = q[i - 1].r + 1; j <= q[i].r; j++)
    {
      if (f[a[j]])
        add(f[a[j]], -1);
      f[a[j]] = j;
      add(j, 1);
    }
    ans[q[i].id] = sum(q[i].r) - sum(q[i].l - 1);
  }
  for (int i = 1; i <= m; i++)
    printf("%d\n", ans[i]);
  return 0;
}