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<bits/stdc++.h>

using namespace std;

#define SpeedForce ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define int long long int
#define endl "\n"
#define loop(i,a,b) for(int i = a; i < b; i++)

const int N = 1e5+5;
int a[N];

int32_t main()
{
	SpeedForce;
	int n;cin >> n;
	loop(i,0,n)
	{
		cin >> a[i];
	}
	sort(a,a+n);
	if(a[n-1] > a[n-2] + a[n-3])
	{
		cout <<"No";
	}
	else
	{
		cout <<"Yes";
	}
	return 0;
}