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.buffer.readline

# 足す時はi番目に足し、返すのは累積和
class sumBIT():
    def __init__(self, N):
        self.N = N
        self.bit = [0 for _ in range(self.N+1)]
    
    def __str__(self):
        ret = []
        for i in range(1, self.N+1):
            ret.append(self.__getitem__(i))
        return "[" + ", ".join([str(a) for a in ret]) + "]"

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

    def add(self, i, x):
        while i <= self.N:
            self.bit[i] += x
            i += i & -i

N, Q = map(int, input().split())
C = list(map(int, input().split()))
LR = [list(map(int, input().split())) for _ in range(Q)]

Toask = [[] for _ in range(N+1)]
for i, (l, r) in enumerate(LR):
    Toask[l].append((r, i))

bit = sumBIT(N+3)
ans = [-1]*Q

Color = [-1]*(N+1)
for l in reversed(range(1, N+1)):
    c = C[l-1]
    if Color[c] != -1:
        last = Color[c]
        bit.add(last, -1)
    bit.add(l, 1)
    for r, ind in Toask[l]:
        ans[ind] = str(bit[r])
    Color[c] = l

print("\n".join(ans))