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>
int main(void){
  int N,i,j,tmp,sum=0;
  scanf("%d",&N);
  int L[N];
  for(i=0;i<N;i++){
    scanf("%d",&L[i]);
  }
  for(i=N-1;i>0;i--){
    for(j=0;j<i;j++){
      if(L[j]>L[j+1]){
      tmp=L[j+1];
      L[j+1]=L[j];
      L[j]=tmp;
      }
    }
  }
  for(i=0;i<N-1;i++){
    sum+=L[i];
  }
  if(sum>L[N-1]){
    printf("Yes");
  }
  else
    printf("No");
  return 0;
}