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,k=map(int,input().split())
d=dict()
lo=hi=ans1=ans2=0
lst=list(map(int,input().split()))
for i in range(n):   
    if lst[i] not in d:
        d[lst[i]]=1
    else:
        d[lst[i]]+=1
    hi+=1                       
    if len (d) > k: 
       
        if ans2-ans1 <= (hi-1)-lo:
            ans2=hi-1 ;ans1=lo
        while len(d) == k+1 :
            d[lst[lo]]-=1
            if d[lst[lo]] == 0:
                del d[lst[lo]]
            lo+=1

                   
if ans2-ans1 <= (hi-1)-lo:
    ans2=hi ;ans1=lo
        
print(ans1+1 , ans2)
