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
import sys
input = sys.stdin.readline

N = int(input())
A = list(map(int, input().split()))

if len(set(A)) == N:
    print(-1)

else:
    S = set()
    res = []
    lt = 1

    for i in range(N):
        if A[i] in S:
            res.append([lt, i + 1])
            S = set()
            lt = i + 2
        else:
            S.add(A[i])
    res[-1][-1] = N
    
    print(len(res))

    for i in range(len(res)):
        res[i] = ' '.join(map(str, res[i]))

    print('\n'.join(map(str, res)))