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;
#define rep(i,n) for(int i=0; i<(n); i++)
constexpr int MX = 500005;
int n, q, c[MX], l, r, p[MX], ans[MX];
struct Q{
	int l, r, i;
} qs[MX];
struct BIT{
	vector<int> dat;
	int sz;
	BIT(int n):sz(n+1), dat(n+1){};
	void set(int x, int n){
		for(int i = n; i < sz; i += i&-i) dat[i] += x;
	}
	int get(int n){
		int ret = 0;
		for(int i = n; i > 0; i -= i&-i) ret += dat[i];
		return ret;
	}
};
int main(){
	scanf("%d%d", &n, &q);
	BIT bt(n);
	rep(i,n) scanf("%d", c+i+1);
	rep(i,q){
		scanf("%d%d", &l, &r);
		qs[i] = Q{l,r,i};
	}
	sort(qs, qs+q, [](Q &a, Q &b){return a.r < b.r;});
	r = 1;
	rep(i,q){
		for(; r <= qs[i].r; r++){
			if(p[c[r]]) bt.set(-1, p[c[r]]);
			bt.set(1, p[c[r]] = r);
		}
		ans[qs[i].i] = bt.get(qs[i].r) - bt.get(qs[i].l-1);
	}
	rep(i,q) printf("%d\n", ans[i]);
}