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 defaultdict

n, k = [int(x) for x in input().split()]
A = [int(x) for x in input().split()]
D = defaultdict(int)
l = 0
argmax = (0, 0)
for i in range(n):
    D[A[i]] += 1
    if len(D) > k:
        if argmax[1] - argmax[0] < i-1 - l: argmax = (l, i-1)
        while len(D) > k:
            D[A[l]] -= 1
            if D[A[l]] == 0: D.pop(A[l])
            l += 1
if argmax[1] - argmax[0] < n-1 - l: argmax = (l, n-1)
print(argmax[0] + 1, argmax[1] + 1)
