human
Write me a Python program to print "yes" or "no" (without quotes), depending on the answer given the following:
Being a programmer, you like arrays a lot. For your birthday, your friends have given you an array a consisting of n distinct integers.

Unfortunately, the size of a is too small. You want a bigger array! Your friends agree to give you a bigger array, but only if you are able to answer the following question correctly: is it possible to sort the array a (in increasing order) by reversing exactly one segment of a? See definitions of segment and reversing in the notes.

Input

The first line of the input contains an integer n (1 ≤ n ≤ 105) — the size of array a.

The second line contains n distinct space-separated integers: a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 109).

Output

Print "yes" or "no" (without quotes), depending on the answer.

If your answer is "yes", then also print two space-separated integers denoting start and end (start must not be greater than end) indices of the segment to be reversed. If there are multiple ways of selecting these indices, print any of them.

Examples

Input

3
3 2 1

Output

yes
1 3

Input

4
2 1 3 4

Output

yes
1 2

Input

4
3 1 2 4

Output

no

Input

2
1 2

Output

yes
1 1

Note

Sample 1. You can reverse the entire array to get [1, 2, 3], which is sorted.

Sample 3. No segment can be reversed such that the array will be sorted.

Definitions

A segment [l, r] of array a is the sequence a[l], a[l + 1], ..., a[r].

If you have an array a of size n and you reverse its segment [l, r], the array will become:

a[1], a[2], ..., a[l - 2], a[l - 1], a[r], a[r - 1], ..., a[l + 1], a[l], a[r + 1], a[r + 2], ..., a[n - 1], a[n].
Solve the task in PYTHON3.
assistant
#Here is Python program to print "yes" or "no" (without quotes), depending on the answer.

import sys
from functools import reduce
from collections import Counter
import time
import datetime
from math import sqrt,gcd

# def time_t():
#     print("Current date and time: " , datetime.datetime.now())
#     print("Current year: ", datetime.date.today().strftime("%Y"))
#     print("Month of year: ", datetime.date.today().strftime("%B"))
#     print("Week number of the year: ", datetime.date.today().strftime("%W"))
#     print("Weekday of the week: ", datetime.date.today().strftime("%w"))
#     print("Day of year: ", datetime.date.today().strftime("%j"))
#     print("Day of the month : ", datetime.date.today().strftime("%d"))
#     print("Day of week: ", datetime.date.today().strftime("%A"))

def ip(): return int(sys.stdin.readline())

def sip(): return sys.stdin.readline()

def mip(): return map(int,sys.stdin.readline().split())

def mips(): return map(str,sys.stdin.readline().split())

def lip(): return list(map(int,sys.stdin.readline().split()))

def matip(n,m):
    lst=[]
    for i in range(n):
        arr = lip()
        lst.append(arr)
    return lst

def factors(n): # find the factors of a number
    return list(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))

def minJumps(arr, n): #to reach from 0 to n-1 in the array in minimum steps
    jumps = [0 for i in range(n)]
    if (n == 0) or (arr[0] == 0):
        return float('inf')
    jumps[0] = 0
    for i in range(1, n):
        jumps[i] = float('inf')
        for j in range(i):
            if (i <= j + arr[j]) and (jumps[j] != float('inf')):
                jumps[i] = min(jumps[i], jumps[j] + 1)
                break
    return jumps[n-1]

def dic(arr): # converting list into dict of count
    return Counter(arr)

def check_prime(n):
    if n<2:
        return False
    for i in range(2,int(n**(0.5))+1,2):
        if n%i==0:
            return False
    return True 

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