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

class BIT:
    def __init__(self, n):
        self.size = n
        self.tree = [0]*(n+1)
 
    def sum(self, i):
        # [0, i) の要素の総和を返す
        s = 0
        while i>0:
            s += self.tree[i]
            i -= i & -i
        return s
    # 0 index を 1 index に変更  転倒数を求めるなら1を足していく
    def add(self, i, x):
        i += 1
        while i <= self.size:
            self.tree[i] += x
            i += i & -i


n,q = map(int,input().split())
c = list(map(int,input().split()))
Q = []
for i in range(q):
    l,r = map(int,input().split())
    Q.append([l,r,i])

Q.sort(key=lambda x: x[1])

bit = BIT(n)
ind = [-1]*(n+1)

ans = [0]*q
now = 0
for l,r,num in Q:
    while now < r:
        x = c[now]
        if ind[x] == -1:
            ind[x] = now
            bit.add(now,1)
        else:
            bit.add(ind[x],-1)
            ind[x] = now
            bit.add(now,1)
        now += 1
    ans[num] = bit.sum(r)-bit.sum(l-1)
for i in ans:
    print(i)

