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 + 100;
int n, Q, a[N], last[N], p[N], ans[N], tree[N];
vector < pair < int, int > > q[N];
void upd(int x, int y) {
    if(!x) return;
    while(x <= 2 * n) {
        tree[x] += y;
        x += x & -x;
    }
} 
int cnt(int x) {
    int ans = 0;
    while(x) {
        ans += tree[x];
        x -= x & -x;
    }
    return ans;
}
int cnt(int l, int r) {
    return cnt(r) - cnt(l - 1);
}
int main() {
    scanf("%d%d", &n, &Q);
    for(int i = 1; i <= n; ++i) {
        scanf("%d", &a[i]);
        a[i + n] = a[i];
    }
    for(int i = 1; i <= 2 * n; ++i) {
        p[i] = last[a[i]];
        last[a[i]] = i; 
    }
    for(int i = 1, l, r; i <= Q; ++i) {
        scanf("%d%d", &l, &r);
        q[r + n].push_back({l + n, i});
    }
    for(int r = 1; r <= 2 * n; ++r) {
        upd(p[r], 1);
        for(auto [l, index] : q[r]) {
            ans[index] = r - l + 1 - cnt(l, r);
        }
    }
    for(int i = 1; i <= Q; ++i) {
        printf("%d\n", ans[i]);
    }
}