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=input()
ar=map(int,raw_input().split())
coun=0
lst=[]
dct={};
flag=1
for i in range(0,n):
    a=ar[i]
    if(flag==1):
        lst.append(i+1)
        flag=0
    if(dct.get(a)==None): dct[a]=1
    else: dct[a]+=1
    if(dct[a]==2):
        dct.clear()
        coun+=1
        flag=1
        lst.append(i+1)
length=len(lst)
if(coun==0):
    print "-1"
else:
    print coun
    if(length%2==1):
        lst.pop()
        lst.pop()
        lst.append(n)
        length-=1
    for i in range(0,length):
        if(i%2==0):
            print lst[i],
        else:
            print lst[i]
