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
class BIT:
    def __init__(self, N):
        self.size = N
        self.tree = [0] * (N + 1)

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

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

N, Q, *CLR = map(int, open(0).read().split())
C, LR = CLR[:N], CLR[N:]

memo = [-1] * (N + 1)
R = []
for i, c in enumerate(C):
    if memo[c] != -1:
        R.append((memo[c], i))
    memo[c] = i

R.sort(key=lambda x: x[0])

LRI = sorted(
    ((l - 1, r - 1, i) for i, (l, r) in enumerate(zip(*[iter(LR)] * 2))),
    key=lambda x: -x[0]
)

bit = BIT(N + 5)
ans = [0] * Q
cur = len(R) - 1
for l, r, i in LRI:
    while 0 <= cur and l <= R[cur][0]:
        bit.add(R[cur][1], 1)
        cur -= 1
    ans[i] = r - l + 1 - bit.sum(r)

for a in ans:
    print(a)