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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <cctype>
#include <cstdlib>
#include <string>
#include <algorithm>
#include <vector>
#include <map>
#define rep(i, n) for(int i = 0; i < n; ++i)
using namespace std;

int comp(const void* a, const void* b){
	return *(int*)a - *(int*)b;
}

int main() {
	int n, l[15], sum = 0;
	cin >> n;
	rep(i, n) cin >> l[i];
	qsort(l, n, sizeof(int), comp);
	for(int i = 1; i < n; i++){
		sum += l[i];
	}
	if(l[0] < sum){
		cout << "Yes" << endl;
	}else{
		cout << "No" << endl;
	}
}