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
import sys
from operator import itemgetter
def input(): return sys.stdin.readline().rstrip()

class BinaryIndexedTree:  # 0-index !!!
    def __init__(self, n):
        self.size = n
        self.tree = [0] * (n + 1)  # 内部的には1-indexで扱う

    def add(self, i, x):  # 0-index
        i += 1  # 内部的には1-indexで扱う
        while i <= self.size:
            self.tree[i] += x
            i += i & -i

    def sum(self, i):  # 0-index  sum of [0,i)
        s = 0
        while i > 0:
            s += self.tree[i]
            i -= i & -i
        return s

    def query(self, l, r):  # 0-index  sum of [l,r)
        return self.sum(r) - self.sum(l)

# ---------------------- #

n, q = (int(x) for x in input().split())
C = list(int(x) - 1 for x in input().split())
LR = [tuple(int(x) for x in input().split()) for _ in range(q)]

bit = BinaryIndexedTree(n)
color_right_index = [-1] * n
queries = [(i, l - 1, r - 1) for i, (l, r) in enumerate(LR)]
queries.sort(key=itemgetter(2))
ANS = [0] * q
k = 0
for i, l, r in queries:
    while k <= r and k <= n:
        color = C[k]
        if color_right_index[color] != -1:
            bit.add(color_right_index[color], -1)
        color_right_index[C[k]] = k
        bit.add(k, 1)
        k += 1
    ANS[i] = bit.query(l, r + 1)

for ans in ANS:
    print(ans)
