Problem: 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

Solution (in PYTHON3):

Solution:
'''
Created on Jan 25, 2016

@author: KANDARP
'''
import sys
n=int(sys.stdin.readline());
arr=[int(x) for x in input().split()];
sets=set();
result=[];
ans=0;
l=0;
r=0;
for i in arr:
    if(i in sets):
        ans=ans+1;
        sets.clear();
        result.append([l+1,r+1]);
        l=r+1;
    else:
        sets.add(i);
    r=r+1;
if ans==0:
    print(-1);
else:
    result[len(result)-1]=[result[len(result)-1][0],n];
    print(ans);    
    print('\n'.join('{0} {1}'.format(p, q) for (p, q) in result));