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=int(input())
ar=list(map(int,input().split()))
s=set()
ans=[]
p=1
for i in range(len(ar)):
    if ar[i] not in s:
        s.add(ar[i])
    else:
        ans.append([p,i+1])
        p=i+2
        s.clear()
if len(ans)!=0 and len(s):
    ans[-1]=[ans[-1][0],len(ar)]
if len(ans):
    print(len(ans))
    for i in ans:
        print(i[0],i[1])
else:
    print(-1)
