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 <bits/stdc++.h>
using namespace std;
const int N = 5e5+1;

int n, p, q, c[N], g[N], h[N], l[N], r[N], ans[N];

bool compare(int i, int j) {
	int u = l[i]/p, v = l[j]/p; return u != v ? (u > v) : (r[i]/p > r[j]/p); }

int main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cin >> n >> q;
    for (int i = 0; i < n; ++i)
    	cin >> c[i];
    for (int i = 0; i < q; ++i)
        cin >> l[i] >> r[i], --l[i], --r[i];
    for (p = 1; p*p <= n; ++p);
    iota(h,h+q,0), --p, sort(h,h+q,compare), ++g[c[0]];
    for (int i, u, v, w = 0, z = 0, j = 0, k = 1; j < q; ++j, ans[i] = k) {
    	for (i = h[j], u = l[i], v = r[i]; z > v; --z)
        	if (--g[c[z]] == 0)
            	--k;
    	for (; z < v; ++g[c[z]])
        	if (g[c[++z]] == 0)
               ++k;
    	for (; w < u; ++w)
        	if (--g[c[w]] == 0)
            	--k;
    	for (; w > u; ++g[c[w]])
        	if (g[c[--w]] == 0)
            	++k; }
    for (int i = 0; i < q; ++i)
        cout << ans[i] << '\n'; }
