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

inputNum=input()
inputTypes=input().split(' ')
    
set1=set()
stack2=[]
stack3=[]
for i in range(len(inputTypes)):
    if len(set1)==0:
        set1.add(inputTypes[i])
        stack2.append(str(i+1))
    else:
        if inputTypes[i] in set1:            
            stack2.append(str(i+1))
            stack3.append(' '.join(stack2))            
            set1.clear()
            stack2.clear()
        else:
            set1.add(inputTypes[i])
if len(stack3)==0:
    print(-1)
    sys.exit()

if len(set1)==0:
    print(len(stack3))
    for x in stack3:
        print(x)
else:
    print(len(stack3))
    lastitem=str(stack3[-1].split()[-1])
    for x in stack3:
        print(x.replace(lastitem,str(len(inputTypes))))
        	 		 		 		 		       		