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 __future__ import nested_scopes, generators, division, absolute_import, with_statement, print_function, unicode_literals
import sys, os, time
input = raw_input
range = xrange

def rv(): return map(int, input().split())
def rl(n): return [list(map(int, input().split())) for _ in range(n)]
if os.path.exists("test.txt"): sys.stdin = open("test.txt")
# sys.stdout = open(pass, 'w')



def solve():
    n, = rv()
    a, = rl(1)
    left = 0
    s = set()
    res = list()
    done = False
    for i in range(n):
        if done:
            s = set()
            done = False
        if a[i] in s:
            res.append([left, i])
            done = True
            left = i + 1
        else:
            s.add(a[i])
    if len(res) == 0:
        print(-1)
    else:
        res[-1][1] = n - 1
        print(len(res))
        print('\n'.join(str(l + 1) + " " + str(r + 1) for l, r in res))

solve()