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
input = sys.stdin.readline


class Bit:
    def __init__(self, n):
        self.size = n
        self.tree = [0] * (n + 1)

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

    def add(self, i, x):
        while i <= self.size:
            self.tree[i] += x
            i += i & -i


N, Q = map(int, input().split())
c = list(map(int, input().split()))

lr = []
for i in range(Q):
    l, r = map(int, input().split())
    lr.append([l, r, i])

lr.sort(key=lambda x: x[1])
# print("lr", lr)
rightest = [-1] * (N + 1)
current_q = 0
bit = Bit(N)
ans = [0] * Q

for i in range(N):
    if rightest[c[i]] != -1:
        bit.add(rightest[c[i]] + 1, -1)
    rightest[c[i]] = i
    bit.add(i + 1, 1)
    # print(bit.tree)

    while current_q < Q and lr[current_q][1] == i + 1:
        ans[lr[current_q][2]] = bit.sum(lr[current_q][1]) - bit.sum(lr[current_q][0] - 1)
        current_q += 1

for i in range(Q):
    print(ans[i])
