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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>

int main()
{
	int n; scanf("%d",&n);
	int l[n]; scanf("%d",&l[0]);
	for (int i = 1; i < n; ++i)
	{
		scanf("%d",&l[i]);
		if(l[i]<l[i-1]){
			int tmp=l[i];
			l[i]=l[i-1];
			l[i-1]=tmp;
		}
	}
	int total=0;
	for (int i = 0; i < n-1; ++i) total+=l[i];
	if(l[n-1]<total) printf("Yes");
	else printf("No");
	return 0;
}