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
from sys import stdin

input = stdin.readline
from collections import defaultdict


class BIT:
    def __init__(self, N):
        self.N = N
        self.arr = [0] * (N + 1)

    def add(self, ind, x=1):
        ind += 1
        while ind <= self.N:
            self.arr[ind] += x
            ind += (ind & -ind)

    def get_sum(self, ind):
        ind += 1
        res = 0
        while ind > 0:
            res += self.arr[ind]
            ind -= (ind & -ind)
        return res


def solve():
    N, Q = map(int, input().split())
    c = list(map(int, input().split()))
    query = [tuple(map(int, inp.split())) for inp in stdin.read().splitlines()]
    prev = [-1] * (N+1)
    nxt = [-1] * (N+1)
    for i, v in enumerate(c):
        if prev[v] != -1:
            nxt[prev[v]] = i
        prev[v] = i
    q = defaultdict(list)
    for i, (l, r) in enumerate(query):
        q[l - 1].append((r - 1, i))
    bit = BIT(N)
    ans = [0] * Q
    for l in range(N - 1, -1, -1):
        if nxt[l] != -1:
            bit.add(nxt[l], 1)
        for r, i in q[l]:
            ans[i] = r - l + 1 - bit.get_sum(r)
    for a in ans:
        print(a)

if __name__ == '__main__':
    solve()
