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 (GCC 9.2.1) (judge verdict: Accepted)

#include<stdio.h>

int main( )
{
    int n , num[500] , i , j , sum= 0 , temp ;
    scanf("%d" , &n);
    
    for( i = 1 ; i <= n ; i++)
        scanf("%d" , &num[i] );
    
    for( i = 1 ; i <= n ; i++)
    {
        for( j = 1 ; j < n ; j++)
        {
            if( num[j] > num[j+1] )
            {
                temp = num[j] ;
                num[j] = num[j+1] ;
                num[j+1] = temp ;
            }
        }
    }
    
    for( i = 1 ; i < n ; i++)
        sum = sum + num[i] ;
    
    if( num[n] < sum)
        printf("Yes\n");
    else 
        printf("No\n");
    
    return 0;
}