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>
int main(void)
{
  int k, j, tmp, a, b;

  int number[10];

  int total;
  scanf("%d", &total);

  for (k=0; k<total; ++k){
    scanf("%d", &number[k]);
  }

  for (k=0; k<total; ++k) {
    for (j=k+1; j<total; ++j) {
      if (number[k] > number[j]) {
        tmp =  number[k];
        number[k] = number[j];
        number[j] = tmp;
      }
    }
  }

  b = total - 1;
  a = number - number[b];

  if (number[0] < a) {
    printf("Yes");
  }else if (number[0] >= a){
    printf("No");
  }
return 0;
}