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 math
import sys
input = sys.stdin.readline
n = int(input())
l = list(map(int,input().split()))
s = set(l)
if len(s) == n:
    print(-1)
    exit()

visited = set()
start = 1
ans = []
for i in range(n):
    x = l[i]
    if x in visited:
        visited = set()
        ans.append((start,i+1))
        start = i+2

    else:
        visited.add(x)

if start <= n:
    st = ans[-1][0]
    en = n
    ans.pop()
    ans.append((st,en))

sys.stdout.write(str(len(ans))+"\n")
for i in ans:
    sys.stdout.write(str(i[0]) + " "+str(i[1])+"\n")