Problem: 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

Solution (in PYTHON3):

Solution:
R = lambda : map(int, input().split())
n = int(input())
a = list(R())

s = set()
l = 0
r = []
for i in range(0,len(a)):
    if a[i] in s:
        r.append((l+1,i+1))
        s = set()
        l = i+1
    else:
        s.add(a[i])

if len(s)==0:
    print(len(r))
    for p in r:
        print(*p)
else:
    if len(r)>0:
        print(len(r))
        r[len(r)-1]=(r[len(r)-1][0],n)
        for p in r:
            print(*p)
    else:
        print(-1)