Problem: 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

Solution (in PYTHON3):

Solution:
'''
      ___           ___                         ___                       ___           ___                         ___
     /\__\         /\  \         _____         /\  \                     /\  \         /\  \                       /\__\
    /:/ _/_        \:\  \       /::\  \        \:\  \       ___         /::\  \       |::\  \         ___         /:/ _/_
   /:/ /\  \        \:\  \     /:/\:\  \        \:\  \     /\__\       /:/\:\__\      |:|:\  \       /\__\       /:/ /\  \
  /:/ /::\  \   ___  \:\  \   /:/  \:\__\   ___ /::\  \   /:/__/      /:/ /:/  /    __|:|\:\  \     /:/  /      /:/ /::\  \
 /:/_/:/\:\__\ /\  \  \:\__\ /:/__/ \:|__| /\  /:/\:\__\ /::\  \     /:/_/:/__/___ /::::|_\:\__\   /:/__/      /:/_/:/\:\__\
 \:\/:/ /:/  / \:\  \ /:/  / \:\  \ /:/  / \:\/:/  \/__/ \/\:\  \__  \:\/:::::/  / \:\~~\  \/__/  /::\  \      \:\/:/ /:/  /
  \::/ /:/  /   \:\  /:/  /   \:\  /:/  /   \::/__/       ~~\:\/\__\  \::/~~/~~~~   \:\  \       /:/\:\  \      \::/ /:/  /
   \/_/:/  /     \:\/:/  /     \:\/:/  /     \:\  \          \::/  /   \:\~~\        \:\  \      \/__\:\  \      \/_/:/  /
     /:/  /       \::/  /       \::/  /       \:\__\         /:/  /     \:\__\        \:\__\          \:\__\       /:/  /
     \/__/         \/__/         \/__/         \/__/         \/__/       \/__/         \/__/           \/__/       \/__/

'''
"""
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"""
import sys
import math
import collections
import operator as op
from collections import deque
from math import gcd, inf, sqrt, pi, cos, sin, ceil, log2, floor, log
from bisect import bisect_right, bisect_left, bisect

# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')

from functools import reduce
from sys import stdin, stdout, setrecursionlimit
setrecursionlimit(2**20)


def ncr(n, r):
    r = min(r, n - r)
    numer = reduce(op.mul, range(n, n - r, -1), 1)
    denom = reduce(op.mul, range(1, r + 1), 1)
    return numer // denom  # or / in Python 2


def prime_factors(n):
    i = 2
    factors = []
    while i * i <= n:
        if n % i:
            i += 1
        else:
            n //= i
            factors.append(i)
    if n > 1:
        factors.append(n)
    return (list(factors))


def isPowerOfTwo(x):
    return (x and (not(x & (x - 1))))

MOD = 1000000007
PMOD = 998244353
N = 10**18 + 1
LOGN = 30
alp = 'abcdefghijklmnopqrstuvwxyz'
T = 1
# T = int(stdin.readline())
for _ in range(T):
    # n, k = list(map(int, stdin.readline().rstrip().split()))
    n = int(stdin.readline())
    a = list(map(int, stdin.readline().rstrip().split()))
    # b = list(map(int, stdin.readline().rstrip().split()))
    # s = str(stdin.readline().strip('\n'))
    # s = list(stdin.readline().strip('\n'))
    # m = int(stdin.readline())
    # c = list(map(int, stdin.readline().rstrip().split()))
    d = {}
    ans = []
    l = 0
    r = 0
    while l < n and r < n:
        if a[r] not in d:
            d[a[r]] = 1
            r += 1
        else:
            d[a[r]] += 1
            if d[a[r]] == 2:
                # while d[a[r]] == 2:
                #     d[a[l]] -= 1
                #     l += 1
                    # print("l:", l, "r:", r, d)
                d = {}
                ans.append([l + 1, r + 1])
                r += 1
                l = r
            else:
                r += 1
        # r += 1
    if len(ans) == 0:
        print(-1)
    else:
        ans[len(ans) - 1][1] = n
        print(len(ans))
        for i in ans:
            print(*i)