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
n,q = map(int, input().split())
cs = list(map(int, input().split()))

prevs = [-1] * (n+1)
# 左端
sts = [-1] * n
for i,c in enumerate(cs):
    c -= 1
    if prevs[c] != -1:
        sts[prevs[c]] = i
    prevs[c] = i

qs = [[] for i in range(n)] 
for i in range(q):
    l,r = map(int, input().split())
    l -= 1
    r -= 1
    qs[l].append((r,i))

class BIT:
    def __init__(self, n):
        self.n = n
        self.bit = [0] * (n+10)

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

    def sum(self, x):
        cnt = 0
        while x > 0:
            cnt += self.bit[x]
            x -= x&-x
        return cnt

bit = BIT(n)
ans = [0] * q
for l in range(n-1,-1,-1):
    # 矢印を追加
    if sts[l] != -1:
        bit.add(sts[l], 1)
    for r,i in qs[l]:
        cnt = bit.sum(r)
        ans[i] = r - l + 1 - cnt

print(*ans, sep="\n")