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 collections import defaultdict
import sys
# Binary Indexed Tree (Fenwick Tree)
class BIT():
    def __init__(self, n):
        '''
        n = 要素数
        要素の添字iは 0 <= i < n となる
        '''
        self.n = n
        self.bit = [0] * (n + 1)

    def add(self, i, value):
        '''i番目の要素にvalueを加算する O(logN)'''
        i = i + 1
        while i <= self.n:
            self.bit[i] += value
            i += i & -i

    def _sum(self, i):
        s = 0
        while i > 0:
            s += self.bit[i]
            i -= i & -i
        return s

    def sum(self, l, r):
        '''区間[l, r)の和を求める O(logN)'''
        return self._sum(r) - self._sum(l)


input = sys.stdin.readline
N, Q = map(int, input().split())
C = list(map(int, input().split()))
LR = [tuple(map(int, input().split())) for _ in range(Q)]
query = [[] for _ in range(N)]
for i, (l, r) in enumerate(LR):
    query[l-1].append((i, r-1))

hist = defaultdict(lambda: -1)
nx = [-1] * N
for i in range(N-1, -1, -1):
    nx[i] = hist[C[i]]
    hist[C[i]] = i

bit = BIT(N)
ans = [None]*Q
for l in range(N-1,-1,-1):
    if nx[l] != -1:
        bit.add(nx[l],-1)
    for i,r in query[l]:
        ans[i] = (r-l+1) + bit.sum(l,r+1)

print(*ans,sep = '\n')