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: Accepted)

#include <iostream>
#include <algorithm>
#include <vector>

using namespace::std;

int serch_max(vector<double> nums) {

	vector<double>::iterator iter = max_element(nums.begin(), nums.end());
	int index = distance(nums.begin(), iter);
	
	return index;
}


int main()
{

	int N;
	cin >> N;

	vector<double> nums(N);

	for (int i = 0; i < N; i++) {
		cin >> nums[i];
	}

	int max_i = serch_max(nums);

	int sum = 0;

	for (int i = 0; i < N; i++) {
		if (i != max_i)
			sum += nums[i];
	}
	
	//cout << sum << endl;
	
	if (nums[max_i] < sum) {
		cout << "Yes" << endl;
	}
	else {
		cout << "No" << endl;
	}


	return 0;
}