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 = 1e6 + 5;
int n, m, a[N], vis[N], t[N];

struct node {
	int l, r, idx, ans;
}b[N];

void upd(int x, int v) {for(; x <= n; x += x & -x) t[x] += v;}
int qry(int x) {int res = 0; for(; x; x -= x & -x) res += t[x]; return res;}
int qry(int l, int r) {return qry(r) - qry(l - 1);}

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", &b[i].l, &b[i].r), b[i].idx = i;
    sort(b + 1, b + m + 1, [](const node &x, const node &y) {
    	return x.r != y.r ? x.r < y.r : x.l < y.l;
    });
    int r = 1;
    for (int i = 1; i <= m; i++) {
        for (int j = r; j <= b[i].r; j++) {
        	if(vis[a[j]]) upd(vis[a[j]], -1);
        	upd(j, 1);
            vis[a[j]] = j;
        }  
        r = b[i].r + 1;
        b[i].ans = qry(b[i].l, b[i].r);
    }
    sort(b + 1, b + m + 1, [](const node &x, const node &y) {
    	return x.idx < y.idx;
    });
    for (int i = 1; i <= m; i++) printf("%d\n", b[i].ans);
    return 0;
}