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
#code
n=int(input())
a=list(map(int,input().split()))
if(len(set(a))==n):
    print(-1)
else:
    ls=[]
    b=set()
    l=1
    b.add(a[0])
    for i in range(1,n):
        if(a[i] not in b):
            b.add(a[i])
        else:
            ls.append([l,i+1])
            l=i+2
            b=set()
    if(ls[-1][1]!=n):
        ls[-1][1]=n
    print(len(ls))
    for i in range(len(ls)):
        print(*ls[i])