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: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

int main(int argc, char *argv[])
{
  int n;
  scanf("%d", &n);
  int *x, i;
  x = (int*)malloc(sizeof(int) * n);
  for(i=0;i<n;i++){
    scanf("%d", x+i);
  }

  int mi;
  mi = 0;
  for(i=1;i<n;i++){
    if(x[i] > x[mi]){
      mi = i;
    }
  }
  /* printf("mi = %d\n", mi); */
  int s=0;
  for(i=0;i<n;i++){
    if(i==mi){
      continue;
    }else{
      s+=x[i];
    }
  }
  /* printf("s = %d\n", s); */

  if(s > x[mi]){
    printf("Yes\n");
  }else{
    printf("No\n");
  }
  free(x);
  return 0;
}