codeforce-B. Sort the Array

B. Sort the Array

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Sample test(s)

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].

判断是否可以通过一次翻转某一个子串使得整个数组有序

保存下来所有值和这个数的id，排序之后与初始的比较，如果只有一段id是逆序的，就证明可以

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
struct node
{
    int id , k ;
} p[100010];
bool cmp(node a,node b)
{
    return a.k < b.k ;
}
int q[100010] ;
int main()
{
    int i , j , n ;
    while(scanf("%d", &n)!=EOF)
    {
        for(i = 0 ; i < n ; i++)
        {
            scanf("%d", &q[i]);
            p[i].id = i ;
            p[i].k = q[i] ;
        }
        sort(p,p+n,cmp);
        for( i = 0 ; i < n ; i++)
            if( p[i].k != q[i] )
                break;
        for(j = n-1 ; j >= 0 ; j--)
            if(p[j].k != q[j] )
                break;
        if(i == n)
            printf("yes\n1 1\n");
        else
        {
            int k = j ;
            for( ; i <= j ; i++)
            {
                if( p[i].id == k )
                    k-- ;
                else
                    break;
            }
            if( i == j+1 )
                printf("yes\n%d %d\n",k+2,  j+1 );
            else
                printf("no\n");
        }
    }
    return 0;
}