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
from collections import *

class BIT:
    def __init__(self, n):
        self.n = n
        self.bit = [0]*(n+1)
    
    def add(self, i, x):
        i += 1
        
        while i<=self.n:
            self.bit[i] += x
            i += i&(-i)

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

N, Q = map(int, input().split())
c = list(map(int, input().split()))
prev = [-1]*N
ys = defaultdict(list)

for i in range(N):
    if prev[c[i]-1]!=-1:
        ys[prev[c[i]-1]].append(i)
    
    prev[c[i]-1] = i

que = defaultdict(list)

for i in range(Q):
    l, r = map(int, input().split())
    que[l-1].append((i, r-1))
    
idx = 0
bit = BIT(N)
ans = [-1]*Q

for x in range(N-1, -1, -1):
    for y in ys[x]:
        bit.add(y, 1)
    
    for i, r in que[x]:
        ans[i] = r-x+1-bit.acc(r+1)

for ans_i in ans:
    print(ans_i)