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

n,Q = map(int,input().split())
c = [0] + list(map(int,input().split()))
dg = 10**6
dg2 = 10**12
res = [0]*Q
LRI = []
for i in range(Q):
    l,r = map(int,input().split())
    LRI.append(l*dg2 + r*dg + i)
LRI.sort()

#A1 ... AnのBIT(1-indexed)
BIT = [0]*(n+1)

#A1 ~ Aiまでの和 O(logN)
def BIT_query(idx):
    res_sum = 0
    while idx > 0:
        res_sum += BIT[idx]
        idx -= idx&(-idx)
    return res_sum

#Ai += x O(logN)
def BIT_update(idx,x):
    while idx <= n:
        BIT[idx] += x
        idx += idx&(-idx)
    return

fi = [[] for i in range(n+1)]
nxt = [-1]*(n+1)
for i in range(1,n+1):
    num = c[i]
    if len(fi[num]) == 0:
        BIT_update(i,1)
    else:
        nxt[fi[num][-1]] = i
    fi[num].append(i)

idx = 0

for i in range(1,n+1):

    while idx < Q:
        l,ri = divmod(LRI[idx],dg2)
        if l != i:
            break
        r,resi = divmod(ri,dg)
        res[resi] = BIT_query(r)
        idx += 1
    BIT_update(i,-1)
    if nxt[i] != -1:
        BIT_update(nxt[i],1)


print(*res,sep='\n')
