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.buffer.readline

N,Q = map(int,input().split())
BIT = [0]*(N+1)

def BIT_query(idx):
    res_sum = 0
    while idx > 0:
        res_sum += BIT[idx]
        idx -= idx&(-idx)
    return res_sum

def BIT_update(idx,x):
    while idx <= N:
        BIT[idx] += x
        idx += idx&(-idx)
    return

c = [0]+list(map(int,input().split()))
lastAppeared = [-1]*(N+1)
ans = [0]*Q
queries = []
for q in range(Q):
    l,r = map(int,input().split())
    queries.append(r*10**12+l*10**6+q)
queries.sort()

curR = 0
for i in range(Q):
    query = queries[i]
    r = query//10**12
    query %= 10**12
    l = query//10**6
    q = query % 10**6
    while curR < r:
        curR += 1
        color = c[curR]
        last = lastAppeared[color]
        if last != -1:
            BIT_update(last,-1)
        lastAppeared[color] = curR
        BIT_update(curR,1)
    ans[q] = BIT_query(r)-BIT_query(l-1)
for i in range(Q):
    print(ans[i])
