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>

int main() {
   int N;
   scanf("%d", &N);

   int L[N];

   for(int i = 0; i < N; i++) {
      scanf("%d", &L[i]);
   }

   int max = L[0];

   for(int i = 0; i < N; i++) {
      if(max < L[i]) {
         max = L[i];
      }
   }

   int sum = 0;

   for(int i = 0; i < N; i++) {
      if(L[i] != max)  {
         sum += L[i];
      }
   }

   if(max < sum) {
      printf("Yes\n");
   } else {
      printf("No");
   }
      
   return 0;
}