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
from sys import stdin

rints = lambda: [int(x) for x in stdin.readline().split()]
n, m = rints()
a, ans, mem, be, dis = rints(), [1, 1], [0] * (10 ** 6 + 1), 0, 0

for i in range(n):
    mem[a[i]] += 1
    if mem[a[i]] == 1:
        dis += 1

    while dis > m:
        mem[a[be]] -= 1
        if not mem[a[be]]:
            dis -= 1
        be += 1
    if i - be > ans[1] - ans[0]:
        ans = [be + 1, i + 1]
print('%d %d' % (ans[0], ans[1]))
