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
def input(): return sys.stdin.readline().strip()
def mapint(): return map(int, input().split())
sys.setrecursionlimit(10**9)

N, Q = mapint()
Cs = list(mapint())
LR = []
for i in range(Q):
    l, r = mapint()
    k = (r<<40)+(l<<20)+i
    LR.append(k)
LR.sort()
mask = (1<<20)-1
A = [0]*N

def add(A,a,w):#リストに値を追加する関数
    x = a
    while x <= N:
        A[x-1] += w
        x += x&(-x)

def sums(A,a):#k番目までの和
    x = a
    S = 0
    while x != 0:
        S += A[x-1]
        x -= x&(-x)
    return S

ans = [0]*Q
i = 0
last_appeared = [-1]*(N+1)
for k in LR:
    ans_idx = k&mask
    l = (k>>20)&mask
    r = (k>>40)
    while i<r:
        c = Cs[i]
        if last_appeared[c]==-1:
            last_appeared[c] = i+1
            add(A, i+1, 1)
        else:
            past = last_appeared[c]
            last_appeared[c] = i+1
            add(A, past, -1)
            add(A, i+1, 1)
        i += 1
    ans[ans_idx] = sums(A, r)-sums(A, l-1)
for a in ans:
    print(a)