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
def longestK(arr,k):
     cnt=[0]*(10**6+1)
     cur=0
     l=0
     ans=0
     al=0
     ar=0
     for i in range(len(arr)):
          cnt[arr[i]]+=1
          if cnt[arr[i]]==1:
               cur+=1
          while cur>k:
               cnt[arr[l]]-=1
               if cnt[arr[l]]==0:
                    cur-=1
               l+=1
          if i-l+1>ans:
               al=l
               ar=i
               ans=i-l+1
     print(al+1,ar+1)
     return ""

a,b=map(int,input().strip().split())
lst=list(map(int,input().strip().split()))
print(longestK(lst,b))