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
from sys import stdin

n = int(stdin.buffer.readline())
a = list(map(int, stdin.buffer.readline().split()))

mp = dict()
l = 0
res = list()
for i in range(n):
    if a[i] in mp:
        res.append((l + 1, i + 1))
        mp.clear()
        l = i + 1
    else:
        mp[a[i]] = 1
if len(res) == 0:
    print(-1)
    exit()
print(len(res))
for i in range(len(res)):
    if i == len(res) - 1:
        print(res[i][0],n)
    else:
        print(res[i][0],res[i][1])
