Gym_100971_Triangles

There is a set of n segments with the lengths l_i. Find a segment with an integer length so that it could form a non-degenerate triangle with any two segments from the set, or tell that such segment doesn't exist.

Input

The first line contains a single integer n (2 ≤ n ≤ 200000) — the number of segments in the set.

The second line contains n integers l_i separated by spaces (1 ≤ l_i ≤ 10⁹) — the lengths of the segments in the set.

Output

If the required segment exists, in the first line output «YES» (without quotes). In this case in the second line output a single integer x — the length of the needed segment. If there are many such segments, output any of them.

If the required segment doesn't exist, output «NO» (without quotes).

Example

Input

23 4

Output

YES2

Input

33 4 8

Output

YES6

Input

33 4 9

Output

NO

题意：给你一组数，问你能否找到一个数使这个数和这组数中任意两个数组成三角形。

解：

      在这组数中挑最小的两组确定一个x的范围，再挑最小和最大的数确定一个x的范围，两个范围一交集就能的到x了。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cstdlib>using namespace std;const int maxn=2e5+5;int a[maxn];int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=0;i<n;i++)        scanf("%d",&a[i]);        sort(a,a+n);        int b,c;        b=a[0]+a[1];        c=a[n-1]-a[0];        if(c+1<b)        printf("YES\n%d\n",c+1);        else        printf("NO\n");    }    return 0;}