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
n=raw_input()
a=map(int,raw_input().split())
index = 0
dic = {}
res = []
for i in range(len(a)):
    if a[i] in dic:
        res.append([index+1,i+1])
        index = i+1
        dic={}
    
    else:
        dic[a[i]]=1

if(len(res)==0):
    print "-1"

else:
    print len(res)
    res[-1][1]=n # -1 means the last element  ->  res[len(res)-1][1]
    for i in res :
        print i[0],i[1]
