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>

int max_index(int array[], int size){
  int p, temp = 0, index;
  
  for(p = 0; p < size; p++){
    if(temp < array[p]){
      temp = array[p];
      index = p;
    }
  }
  
  return index;
}

int main(void){
  int n, *l;
  int i, j, index, sum = 0;
  
  scanf("%d", &n);
  
  l = (int *)malloc(n * sizeof(int));
  
  for(i = 0; i < n; i++){
    scanf("%d", &l[i]);
  }
  
  index = max_index(l, n);
  
  for(j = 0; j < n; j++){
    if(j == index){
      continue;
    }
    sum = sum + l[j];
  }
  
  if(sum > l[index]){
    printf("Yes");
  } else {
    printf("No");
  }
  
  return 0;
}