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
from math import sqrt as s
from collections import *
from fractions import gcd
from sys import stdin, stdout
n,k=map(int,raw_input().split())
lst=map(int,raw_input().split())
cnt=[0]*(max(lst)+2)
ans=-1
i=0
j=0
ans=[]
cu=0
ma=-1
while i<n:
    cnt[lst[i]]+=1
    if cnt[lst[i]]==1:
        cu=cu+1
    i=i+1
    while cu>k:
        cnt[lst[j]]-=1
        if cnt[lst[j]]==0:
            cu=cu-1
        j=j+1
    if cu<=k:
        if i-j>ma:
            ma=i-j
            ans=(j,i)
        #ans.append((j,i))
print ans[0]+1,ans[1]
def checkunique(lst,i,j):
    s=set()
    for a in range(i-1,j):
        s.add(lst[a])
    return len(s)
#print checkunique(lst,ans[0]+1,ans[1])








