codeforces 451B Sort the Array

B. Sort the Array

time limit per test

memory limit per test

input

output

a consisting of n distinct

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

n (1 ≤ n ≤ 105) — the size of array a.

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

Output

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

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

[1, 2, 3], which is sorted.

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

Definitions

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

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].

题意：逆转数组的一段，看能不能使它从小到大，如果能，输出这一段的始末位置，如果本身就是从小到大的，输出1 1。注意格式。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
const int ma=1e5+10;
typedef long long ll;
using namespace std;
int a[ma],b[ma];
int main()
{
    int n;
    cin>>n;
    for(int i=1; i<=n; i++)
    {
        cin>>a[i];
        b[i]=a[i];
    }
    sort(b+1,b+n+1);
    int s=0,st=-1,en=-1;
    for(int i=1; i<=n; i++)
    {
        if(a[i]!=b[i])
        {
            s++;
            if(st==-1)
                st=i;
            else
                en=i;
        }
    }
    if(s==0)
    {
        cout<<"yes"<<endl;
        cout<<"1 1"<<endl;
        return 0;
    }
    s=en-st+1;
    int w1=st,w2=en;
    for(int i=1; i<=s/2; i++)
    {
        int temp=a[st];
        a[st]=a[en];
        a[en]=temp;
        //cout<<st<<' '<<en<<endl;
        st++;
        en--;
    }
    for(int i=1; i<=n; i++)
    {
        if(a[i]!=b[i])
        {
            cout<<"no"<<endl;
            return 0;
        }
    }
    cout<<"yes"<<endl<<w1<<" "<<w2<<endl;
    /*for(int i=1;i<=n;i++)
        cout<<a[i]<<' ';
    cout<<endl;
    for(int i=1;i<=n;i++)
        cout<<b[i]<<' ';
    cout<<endl;*/
    return 0;
}