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: Wrong Answer)

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

int max(int a, int b){
  return a >= b ? a : b;
}
int min(int a, int b){
  return b >= a ? a : b;
}
int factorial(int n) {
    if (n > 0) return n*factorial(n - 1);
		else return 1;
}
int compare_up_int(const void *a, const void *b){
   return *(int*)a - *(int*)b;
}
int compare_down_int(const void *a, const void *b){
   return *(int*)b - *(int*)a;
}
int euclid(int a, int b){
  if(a < b){int tmp = a;a = b;b = tmp;}
  int r = a % b;
  if(r < 0) r += b;
  while(r != 0){
    a = b;b = r;r = a % b;
    if(r < 0) r += b;
  }
  return b;
}

int main(){
  int N,count=0;
  scanf("%d",&N);
  int num[N];
  for(int i=0;i<N;i++) scanf("%d ",&num[i]);
  qsort(num,N,sizeof(int),compare_up_int);
  for(int i=0;i<N-1;i++) count += num[i];
  if(count > num[N]){
    printf("Yes");
  }
  if(count < num[N]){
    printf("No");
  }
  if(count == num[N]){
    printf("No");
  }
  return 0;
}