The array a with n integers is given. Let's call the sequence of one or more consecutive elements in a segment. Also let's call the segment k-good if it contains no more than k different values.

Find any longest k-good segment.

As the input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

Input

The first line contains two integers n, k (1 ≤ k ≤ n ≤ 5·105) — the number of elements in a and the parameter k.

The second line contains n integers ai (0 ≤ ai ≤ 106) — the elements of the array a.

Output

Print two integers l, r (1 ≤ l ≤ r ≤ n) — the index of the left and the index of the right ends of some k-good longest segment. If there are several longest segments you can print any of them. The elements in a are numbered from 1 to n from left to right.

Examples

Input

5 5
1 2 3 4 5


Output

1 5


Input

9 3
6 5 1 2 3 2 1 4 5


Output

3 7


Input

3 1
1 2 3


Output

1 1
from collections import *
from math import *
from itertools import *
frq = defaultdict(int)
def I(): return map(int, input().strip().split())
n, k = I()
arr = [0] + list(I())
ans, pivot, nums = [0, 0], 0, 0
l, r = 1, 1
while r <= n:
    if not bool(frq[arr[r]]):
        nums += 1
    frq[arr[r]] += 1
    while nums > k:
        frq[arr[l]] -= 1
        if not bool(frq[arr[l]]):
            nums -= 1
        l += 1
    if (r-l+1) > pivot:
        pivot = r-l+1
        ans = [l, r]
    r += 1
print(*ans)
