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, stdout, exit
n = int(stdin.readline())
a = [int(i) for i in stdin.readline().split()]
ans = []
s = set()
i = 0
j = -1
while True:
    j += 1
    if j == n:
        if len(ans) != 0:
            ans[-1] = (ans[-1][0], j - 1)
        break
    if a[j] in s:
        s = set()
        ans.append((i, j))
        i = j + 1
        j = i - 1
    else:
        s.add(a[j])
if len(ans) == 0:
    print(-1)
    exit(0)
stdout.write(str(len(ans)) + '\n')
for x, y in ans:
    stdout.write(str(x + 1) + ' ' + str(y + 1) + '\n')
    
