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 numpy as np
from sys import exit, stdin
import copy
from collections import deque, Counter


class BIT:
    def __init__(self, a=0):
        self.n = a
        self.d = [0] * (a + 1)

    def add(self, i, x=1):
        i += 1
        while i <= self.n:
            self.d[i] += x
            i += i & (-i)

    def sum(self, *i):
        if len(i) == 1:
            x = 0
            tmp = i[0] + 1
            while tmp:
                x += self.d[tmp]
                tmp -= tmp & (-tmp)
            return x
        else:
            return sum(self, i[1] - 1) - sum(self, i[0] - 1)


N, Q = map(int, stdin.readline().split())
C = list(map(int, stdin.readline().split()))

app = [-1] * (N + 1)
ps = [[] for _ in range(N)]
for i in range(N):
    tmp = app[C[i]]
    if tmp != -1:
        ps[tmp].append(i)
    app[C[i]] = i

qs = [[] for _ in range(N)]
for i in range(Q):
    l, r = map(int, stdin.readline().split())
    l -= 1
    r -= 1
    qs[l].append(copy.deepcopy([r, i]))

ans = [0] * Q
d = BIT(N)
for x in range(N - 1, -1, -1):
    for y in ps[x]:
        d.add(y, 1)
    for p in qs[x]:
        r = p[0]
        i = p[1]
        ans[i] = (r - x + 1) - d.sum(r)

for re in ans:
    print(re)
