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>
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
using namespace std;

typedef long long ll;

const int MAX_N = 5e5;

int bit[MAX_N + 1], n;

int sum(int i) {
    int s = 0;
    while (i > 0) {
        s += bit[i];
        i -= i & -i;
    }
    return s;
}

void add(int i, int x) {
    while (i <= n) {
        bit[i] += x;
        i += i & -i;
        //cout << i << " ";
    }
    //cout << endl;
}

int main() {
    int Q;
    cin >> n >> Q;
    vector<int> c(n), b(n, -1), l(Q), r(Q);
    vector<vector<int>> ls(n);
    map<pair<int, int>, int> ans;
    rep(i, n) {
        cin >> c[i];
        c[i]--;
    }
    rep(i, Q) {
        cin >> l[i] >> r[i];
        l[i]--;
        r[i]--;
        ls[r[i]].push_back(l[i]);
    }
    rep(i, n) {
        add(i + 1, 1);
        if (b[c[i]] > -1) add(b[c[i]] + 1, -1);
        b[c[i]] = i;
        rep(j, ls[i].size()) {
            ans[make_pair(ls[i][j], i)] = sum(i + 1) - sum(ls[i][j]);
        }
    }
    rep(i, Q) {
        cout << ans[make_pair(l[i], r[i])] << "\n";
    }
}