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
n,q= map(int,input().split())
A = list(map(int,input().split()))
C = [list(map(int,input().split())) + [i] for i in range(q)]
for i in range(q):
    C[i][0] -= 1
    C[i][1] -= 1
for i in range(n):
    A[i] -= 1

class Bit:
    def __init__(self, n):
        self.size = n
        self.tree = [0] * (n + 1)
  
    def sum(self, i):
        s = 0
        while i > 0:
            s += self.tree[i]
            i -= i & -i
        return s
  
    def add(self, i, x):
        while i <= self.size:
            self.tree[i] += x
            i += i & -i

B = Bit(n+1)
ANS = [-1] * q
C.sort(key=lambda x:x[1])
G = [-1] * (max(A)+1)
ind = 0
for i in range(n):
    a = A[i]
    if G[a] != -1:
        # print("ADD", G[a]+1, -1)
        B.add(G[a]+1, -1)
    G[a] = i
    # print("ADD", G[a]+1, 1)
    B.add(G[a]+1, 1)
    while ind < q and C[ind][1] == i:
        # print("SUM", C[ind][1]+1, C[ind][0]+1-1)
        ANS[C[ind][2]] = B.sum(C[ind][1]+1) - B.sum(C[ind][0]+1-1)
        ind += 1

for i in range(q):
    print(ANS[i])