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 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<limits.h>
int main()
{
  int *x,n,i,a,sum;
  scanf("%d",&n);
  x = (int*)malloc(sizeof(int) * n);
  for(i=0;i < n;i++){
    scanf("%d",x+i);
  }
  
  for(i=0;i <= n-2;i++){
    if (x[i] > x[i+1]){
      a = x[i+1];
      x[i+1] = x[i];
      x[i] = a;
    }
  }
  sum = 0;
  for (i=0;i < n-1;i++){
    sum += x[i];
  }
  if (sum > x[n-1]){
    printf("Yes\n");
  }else{
    printf("No\n");
  }
  printf("%d\n%d\n",n,x[4]);
  
  return 0;
}