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, k = map(int, sys.stdin.buffer.readline().decode('utf-8').split())
a = list(map(int, sys.stdin.buffer.readline().decode('utf-8').split()))
ans_len, i, j = 0, 0, 0
ans_i, ans_j = 0, 0
kind = 0
cnt = [0]*(10**6+10)

for i in range(n):
    while j < n and kind + (1 if cnt[a[j]] == 0 else 0) <= k:
        kind += 1 if cnt[a[j]] == 0 else 0
        cnt[a[j]] += 1
        j += 1

    if ans_len < j-i:
        ans_len = j-i
        ans_i, ans_j = i+1, j

    cnt[a[i]] -= 1
    kind -= 1 if cnt[a[i]] == 0 else 0

print(ans_i, ans_j)
