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 bitadd(a,w,bit): #aにwを加える(1-origin)
 
    x = a
    while x <= (len(bit)-1):
        bit[x] += w
        x += x & (-1 * x)
 
def bitsum(a,bit): #ind 1～aまでの和を求める
 
    ret = 0
    x = a
    while x > 0:
        ret += bit[x]
        x -= x & (-1 * x)
    return ret

from sys import stdin
N,Q = map(int,stdin.readline().split())

c = list(map(int,stdin.readline().split()))
last = [1] * (N+1)

li = [ [] for i in range(N+2)]
ans = [None] * Q
for i in range(Q):
    L,R = map(int,stdin.readline().split())
    L += 1
    R += 1
    li[R].append((L,i))

BIT = [0] * (N+2)
bitadd(1,N,BIT)

for i in range(N):

    nc = c[i]
    lastind = last[nc]
    last[nc] = i+2

    bitadd(lastind,-1,BIT)
    bitadd(i+2,1,BIT)

    for l,qind in li[i+2]:
        ans[qind] = bitsum(i+2,BIT) - bitsum(l-1,BIT)

print (*ans)

