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, res = int(input()), []
    s, i, fmt = set(), 1, "{:n} {:n}".format
    for j, a in enumerate(input().split(), 1):
        if a in s:
            s = set()
            res.append(fmt(i, j))
            i = j + 1
        else:
            s.add(a)
    if res:
        print(len(res))
        res[-1] = res[-1].split()[0] + ' ' + str(n)
        print('\n'.join(res))
    else:
        print(-1)


if __name__ == '__main__':
    main()
