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 = int(raw_input())

a = []
a = map(int , raw_input().split())

dic = {a[0]:0}

c = 0

p = a[0]
pair = []
k=0
for i in range (1,n):
    if a[i] in dic:
        pair.append((dic[a[i]]+1,i+1))
        dic.clear()
        c+=1
        k=i+1
    else:
        dic[a[i]]=k

if c>0:
   print c
   for i in range(0,c-1):
       (x,y)=pair[i]
       print `x` +" "+ `y`
       
   (x,y) = pair[c-1]
   print `x` +" "+ `n`
else:
    print'-1'
