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
#pragma gcc optimize("Ofast")
#include "bits/stdc++.h"
using namespace std;

int main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n, q; cin >> n >> q;
    vector<int> c(n), la(n, -1), ans(q), S(n << 1);
    for (int i = 0; i < n; ++i) cin >> c[i], --c[i];
    vector<vector<pair<int, int>>> Q(n);
    for (int i = 0; i < q; ++i) {
        int l, r; cin >> l >> r, --l, --r;
        Q[r].emplace_back(l, i);
    }
    auto upd = [&] (int i, int v) {
        for (S[i += n] = v; i >>= 1;) S[i] = S[i << 1] + S[i << 1 | 1];
    };
    auto query = [&] (int l, int r) {
        int res = 0;
        for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
            if (l & 1) res += S[l++];
            if (r & 1) res += S[--r];
        }
        return res;
    };
    for (int i = 0; i < n; ++i) {
        if (la[c[i]] > -1) upd(la[c[i]], 0);
        la[c[i]] = i;
        upd(i, 1);
        for (auto& o : Q[i]) {
            auto& [l, id] = o;
            ans[id] = query(l, i + 1);
        }
    }
    for (int i = 0; i < q; ++i) cout << ans[i] << '\n';
}