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: Wrong Answer)

#include<stdio.h>
int main()
{
   int N,i,j,sum = 0;

   scanf("%d",&N);

   if(N >= 3 && N <= 10)
   {
     int A[N];

     for(i = 0; i < N; i++)
     {
       scanf("%d",&A[i]);

     }

     if(1<=A[i]<=100)
     {
     int max = A[0];

     for(i = 1; i < N; i++)
     {
         if(max < A[i])
              max = A[i];

     }

     for(j = 0; j < N; j++)
     {

       if(A[j] == max)

         continue;

        sum = sum + A[j];

      }

      if(sum > max)
         printf("Yes\n");

      else
        printf("No\n");



     }


   }















 return 0;
}