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;

#define ll long long
#define mp make_pair
#define pb push_back

#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#define rep(i, n) for (int i = 0; i < (int)(n); ++ i)

const int mxn = 5e5 + 5;
int n, q, a[mxn], lst[mxn], ans[mxn];
vector <pair <int, int> > qry[mxn];

struct BIT {
	int s[mxn];
	void upd(int x, int v) { for (++ x; x < mxn; x += x & -x) s[x] += v; }
	int qry(int x) { int ans = 0; for (++ x; x; x -= x & -x) ans += s[x]; return ans; }
} bit;

int main() {
	scanf("%d %d", &n, &q);
	rep(i, n) scanf("%d", &a[i]), -- a[i];
	rep(i, q) {
		int l, r;
		scanf("%d %d", &l, &r);
		-- l; -- r;
		qry[l].pb(mp(r, i));
	}
	rep(i, n) lst[i] = n;
	for (int i = n - 1; ~i; -- i) {
		bit.upd(lst[a[i]], -1); bit.upd(lst[a[i]] = i, +1);
		rep(j, qry[i].size()) ans[qry[i][j].second] = bit.qry(qry[i][j].first);
	}
	rep(i, q) printf("%d\n", ans[i]);
	return 0;
}