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: Wrong Answer)

#include<stdio.h>

int main(void)
{

	int n;
	int a[11];
	int i;
	int x = 0;
	int y = 0;
	scanf("%d", &n);
	for (i = 0; i < n; i++) {
		scanf("%d", &a[i]);

	}
	x = a[0];

	for (i = 0; i < n; i++) {
		if (a[i] < a[i + 1]) {
			x = a[i + 1];

		}

	}
	for (i = 0; i < n; i++) {
		y += a[i];

	}
	if (x < y - x) {
		printf("Yes");

	}
	else {
		printf("No");

	}

	return 0;

}