Codeforces 451B

Being a programmer, you like arrays a lot. For your birthday, your friends have given you an arraya consisting of ndistinct 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 arraya (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 arraya.

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

33 2 1

Output

yes1 3

Input

42 1 3 4

Output

yes1 2

Input

43 1 2 4

Output

no

Input

21 2

Output

yes1 1

这道题用模拟就可以了，直接上代码了。

#include <cstdio>#include <algorithm>using namespace std;const int maxn = 100010;struct Node{    int val,id;    bool operator < (const Node& rhs) const{        return val < rhs.val;    }}node[maxn];int main(){    int n,l = -1,r;    bool flag = 1;    scanf("%d",&n);    for(int i = 1;i <= n;i++){        scanf("%d",&node[i].val);        node[i].id = i;    }    sort(node+1,node+n+1);    node[n+1].id = -1;    for(int i = 1;i <= n;i++){        if(node[i].id != i){            l = i;            break;        }    }    if(l == -1){        printf("yes\n1 1\n");        return 0;    }    for(int i = l;i <= n;i++){        if(node[i].id - node[i+1].id != 1){            r = i;            break;        }    }    for(int i = r+1;i <= n;i++){        if(node[i].id != i){            flag = false;            break;        }    }    if(flag)    printf("yes\n%d %d\n",l,r);    else        printf("no\n");    return 0;}