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

def main():
    n, k = map(int,input().split())
    element_count = [0]*1000001

    count=0
    a = list(map(int, input().split()))
    s = 0
    res=0
    for t in range(n):
        element_count[a[t]] += 1
        if element_count[a[t]] == 1: count += 1
        while count > k:
            element_count[a[s]] -= 1
            if element_count[a[s]] == 0: count -= 1
            s +=1
        if t - s + 1 > res:
            res = t - s + 1
            l = s
            r = t

    print(l+1,r+1)


if __name__ == '__main__':
    main()