Problem: 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

Solution (in PYTHON):

Solution:
def main():
    N = input()
    pears = map(int, raw_input().split())

    counts = {}
    result = []
    left = 1

    
    for i in range(len(pears)):
        if(not counts.has_key(pears[i])):
            counts[pears[i]] = 1
        else:
            counts.clear()
            result.append([left, i+1])
            left = i+2

    if(len(result)):
        result[len(result)-1][1] = N
        

    if(not len(result)):
       print -1

    else:
        print len(result)
        for val in result:
            print val[0], val[1]
    
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if(__name__ == "__main__"):
    main()