Problem:
Score : 200 points

Problem Statement

Determine if an N-sided polygon (not necessarily convex) with sides of length L_1, L_2, ..., L_N can be drawn in a two-dimensional plane.

You can use the following theorem:

Theorem: an N-sided polygon satisfying the condition can be drawn if and only if the longest side is strictly shorter than the sum of the lengths of the other N-1 sides.

Constraints

All values in input are integers.

3 \leq N \leq 10

1 \leq L_i \leq 100

Input

Input is given from Standard Input in the following format:

N
L_1 L_2 ... L_N

Output

If an N-sided polygon satisfying the condition can be drawn, print Yes; otherwise, print No.

Sample Input 1

4
3 8 5 1

Sample Output 1

Yes

Since 8 < 9 = 3 + 5 + 1, it follows from the theorem that such a polygon can be drawn on a plane.

Sample Input 2

4
3 8 4 1

Sample Output 2

No

Since 8 \geq 8 = 3 + 4 + 1, it follows from the theorem that such a polygon cannot be drawn on a plane.

Sample Input 3

10
1 8 10 5 8 12 34 100 11 3

Sample Output 3

No

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
using namespace std;

int main(){
    int n;
    cin >> n;
     int l[n]={};

    for(int i=0;i<n;i++){
        cin >> l[i];
    }

    int a;

    for(int j=0;j<n;j++){
        if(l[j]>l[j+1]){
            a=l[j];
            l[j]=l[j+1];
            l[j+1]=a;
        }
    }

    int sum=0;

    for(int z=0;z<n-1;z++){
        sum=sum+l[z];
    }

   

    if(sum>l[n]){
        cout << "Yes"<< endl;
    }else{
        cout << "No" << endl;
    }

    return 0;
}