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
def main():
    N, Q = map( int, input().split())
    C = list( map( int, input().split()))
    LRI = []
    for i in range(Q):
        l, r = map( int, input().split())
        LRI.append((r,l,i))
    LRI.sort(key=lambda x:x[0])
    lastAppend = [-1]*(N+1)
    BIT = [0]*(N+1)
    #A1 ~ Aiまでの和 O(logN)
    def BIT_query(idx):
        res_sum = 0
        if idx == 0:
            return 0
        while idx > 0:
            res_sum += BIT[idx]
            idx -= idx&(-idx)
        return res_sum

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

    ANS = [0]*Q
    now = 1
    for r, l, i in LRI:
        while now <= r:
            c = C[now-1]
            if lastAppend[c] == -1:
                BIT_update(now, 1)
            else:
                BIT_update( now, 1)
                BIT_update(lastAppend[c],-1)
            lastAppend[c] = now                
            now += 1
        ANS[i] = BIT_query(r) - BIT_query(l-1)
    print( "\n".join( map( str, ANS)))
    
    

if __name__ == '__main__':
    main()
