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
# -*- coding: utf-8 -*-
# @Time    : 2021/2/1 3:39 下午
# @Author  : qu
# @Email   : quzhenqing@zju.edu.cn
# @File    : C. Pearls in a Row.py
from sys import stdin

n = int(stdin.readline().strip())
a = list(map(int, stdin.readline().split()))
s = set()
flag = False
index = 1
res1 = []
res2 = []

for i, v in enumerate(a):
    if not (v in s):
        s.add(v)
    else:
        s.clear()
        res1.append(index)
        res2.append(i + 1)
        index = i + 2
        flag = True
# print(n, a, flag)
if flag:
    if not res2[-1] == n:
        res2[-1] = n
    print(len(res1))
    for i, j in zip(res1, res2):
        print(i, j)
else:
    print(-1)
