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: Accepted)

#include <stdio.h>

/*
「定理 : 一番長い辺が他の N-1 辺の長さの合計よりも真に短い場合に限り、条件を満たす N 角形が描ける。」

からといって、

一番長い辺が他の N-1 辺の長さの合計よりも真に短ければ必ず条件を満たすN角形が描けるとは言えないにゃーん。

(もとの定理の裏)

とはいえ、図形を考えてみれば多分描けるやろ(雑)
*/

int main(void) {
	int N;
	int L[16];
	int i;
	int asumisu = 0, mizuhasu = 0;
	if (scanf("%d", &N) != 1) return 1;
	for (i = 0; i < N; i++) {
		if (scanf("%d", &L[i]) != 1) return 1;
		asumisu += L[i];
		if (mizuhasu < L[i]) mizuhasu = L[i];
	}
	puts(asumisu - mizuhasu > mizuhasu ? "Yes" : "No");
	return 0;
}