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 + 100;
inline int read() {
	int q=0,w=1;char c=getchar();
	while(!isdigit(c)){if(c=='-')w=-1;c=getchar();}
	while(isdigit(c))q=q*10+c-'0',c=getchar();
	return w*q;
}

struct node {
	int l, r, id, bel;
	inline bool operator < (const node &b) const {
		if(bel == b.bel) return r < b.r;
		return l < b.l;
	}
}Q[N];
int n, m, block, col[N], cnt[N], Ans[N];

int main() {
	n = read(); m = read(); block = sqrt(n);
	for(int i = 1;i <= n; ++i) col[i] = read();
	for(int i = 1;i <= m; ++i) {
		Q[i].l = read();
		Q[i].r = read();
		Q[i].id = i;
		Q[i].bel = (Q[i].l + 1) / block;
	}
	sort(Q + 1, Q + 1 + m);
	int l = 1, r = 0, ans = 0;
	for(int i = 1;i <= m; ++i) {
		int ql = Q[i].l, qr = Q[i].r;
        while(l < ql) ans -= !--cnt[col[l++]];
        while(l > ql) ans += !cnt[col[--l]]++;
        while(r < qr) ans += !cnt[col[++r]]++;
        while(r > qr) ans -= !--cnt[col[r--]];
        Ans[Q[i].id] = ans;
	}
	for(int i = 1;i <= m; ++i) 
		printf("%d\n", Ans[i]);
	return 0;
}