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
from operator import itemgetter
N,Q = map(int, input().split())
C = [int(c) for c in input().split()]
L = [[int(l) for l in input().split()] for _ in range(Q)]

for i in range(Q):
    L[i].append(i)
M = pow(2, len(bin(N))-2)
BIT = [0]*(M+1)
L.sort(key=itemgetter(1))
memo = [0]*(N+1)

ans = [0]*Q
i = 1
j = 0
while i <= N:
    c = C[i-1]
    idx = i
    if i <= L[j][1]:
        if memo[c] > 0:
            k = memo[c]
            while k <= M:
                BIT[k] -= 1
                k += k&(-k)
        memo[c] = i
        while idx <= M:
            BIT[idx] += 1
            idx += idx&(-idx)
    if i == L[j][1]:
        cnt = 0
        n = L[j][1]
        while n > 0:
            cnt += BIT[n]
            n -= n&(-n)
        n = L[j][0]-1
        while n > 0:
            cnt -= BIT[n]
            n -= n&(-n)
        ans[L[j][2]] = cnt
        j += 1
        if j == Q:
            break
        elif L[j][1] == i:
            i -= 1
    i += 1

for a in ans:
    print(a)