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
import sys
sys.setrecursionlimit(10**6)
def get_array(): return list(map(int, sys.stdin.readline().split()))
def get_ints(): return map(int, sys.stdin.readline().split())
def input(): return sys.stdin.readline().strip('\n')

n = int(input())
l = get_array()
n = len(l)
ans = []
i , j = 0 , 0
while i < n:
    s = set()
    while j < n and l[j] not in s:
        #print(j)
        s.add(l[j])
        j += 1
    if j == n:
        break

    ans.append([i,j])
    j+= 1
    i = j

if len(ans) == 0:
    print(-1)
else:
    print(len(ans))
    ans[-1][-1] = max(n-1,ans[-1][-1])
    for i in ans:
        print(i[0]+1,i[1]+1)
