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 9.2.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>

typedef long long ll;

int max(int a, int b){
  if(a > b) return a;
  else return b;
}
int min(int a, int b){
  if(a < b) return a;
  else return b;
}

int sort(const void *a ,const void *b){
  return *(int*)b - *(int*)a;
}

int main(){

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

  int l[n];
  for(int i = 0; i < n; ++i){
    scanf("%d",l+i);
  }

  qsort(l,n,sizeof(int),sort);
  int sum = 0;
  for(int i = 1; i < n; ++i){
    sum += l[i];
  }

  if(l[0] < sum) printf("Yes");
  else printf("No");

  return 0;
}