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;
  int L[10] = {0};
  int max = 0;
  int max_num = 0;
  scanf("%d", &N);
  
  for(int i = 0 ; i < N ; i++){
    scanf("%d" , &L[i]);
    
    if(L[i] > max){              //最大値とその場所を記憶
      max = L[i];
      max_num = i;
    }
    
  }


  int sum = 0;
  for(int i = 0 ; i < N ; i++){
    if(i == max_num) continue;   //最大数の場所なら足さなくていい

    sum += L[i];
  }

  if(sum > max){
    puts("Yes");
  }else{
    puts("No");
  }    
  
  return 0;
}