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

n = int(sys.stdin.readline())

ans = []
s = set()
left = 1

for i, e in enumerate(map(int, sys.stdin.readline().split()), 1):
    if e in s:
        s = set()
        ans += [(left, i)]
        left = i + 1
    else:
        s.add(e)
if ans:
    sys.stdout.write(f"{len(ans)}\n")
    ans[-1] = ans[-1][0], n
    for a, b in ans:
        sys.stdout.write(f"{a} {b}\n")
else:
    sys.stdout.write("-1\n")
