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
input=sys.stdin.readline
from collections import defaultdict
n,k=map(int,input().split())
a=list(map(int,input().split()))
cnt=defaultdict(int)
typ=0
d=0
ans=[]
l,r=0,0 # [l,r)
while r<n:
  if typ<=k:
    r+=1
    aa=a[r-1]
    if cnt[aa]==0:
      cnt[aa]=1
      typ+=1
    else:
      cnt[aa]+=1
    if typ<=k:
      if d<r-l:
        d=r-l
        ans=[l+1,r]
  else:
    aa=a[l]
    if r-l>1:
      l+=1
    else:
      break
    if cnt[aa]==1:
      cnt[aa]-=1
      typ-=1
    else:
      cnt[aa]-=1
    if typ<=k:
      if d<r-l:
        d=r-l
        ans=[l+1,r]
print(*ans)