Codeforces Round #258 (Div. 2) B. Sort the Array

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

题目大意：

一个序列求能否够反转某一部分。使得的序列为递增的，注意一句话then also print two space-separated integers denoting start and end。输出的是位置，而不是当前

的数值。

并若是反转有reverse（）函数，用sort()WA。。

思路：从左到右找到比前一个数小的数的下标，然后从右向左找比后一个数大的下标。找到便break

#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
#include<cmath>
#include<algorithm>
#define LL  int
#define inf 0x3f3f3f3f
using namespace std;
LL  a[1000001],b[1000001];
int main()
{
    LL  n,m,i,j,k,x,y;
    while(scanf("%d",&n)!=EOF)
    {
        x=y=1;
        for(i=1; i<=n; i++)
            scanf("%d",&a[i]);
        for(i=1; i<n; i++)
        {
            if(a[i]>a[i+1])
            {
                x=i;
                break;
            }
        }
        for(i=n; i>=2; i--)
            if(a[i]<a[i-1])
            {
                y=i;
                break;
            }
        reverse(a+x,a+y+1);
        bool vis=false;
        for(i=1; i<=n-1; i++)
        {
            if(a[i]>a[i+1])
            {
                vis=true;
            }
        }
        if(vis)
            printf("no\n");
        else
            printf("yes\n%d %d\n",x,y);
    }
    return 0;
}