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())
lis=list(map(int,input().split()))

ins=[0 for i in range(10**6)]
ins[lis[0]]+=1
zeroes=10**6-1
mL=0
pair=[0,0]
r=0
for l in range(n):

    if l>0:
        ins[lis[l-1]]-=1
        if ins[lis[l-1]]==0:
            zeroes+=1

    while 10**6-zeroes<=k and r<n-1:
        r+=1
        ins[lis[r]]+=1
        if ins[lis[r]]==1:
            zeroes-=1

        if 10**6-zeroes==k+1:
            ins[lis[r]]-=1
            zeroes+=1
            r-=1
            break

    if mL<r-l:
        mL=r-l
        pair=[l,r]

print(pair[0]+1,pair[1]+1)
