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<stdio.h>
#include<algorithm>
#include<vector>
#include<string.h>
#include<math.h>
using namespace std;

int syc(long i,long Min,long Max){
	if(Min <= i && i <= Max){
		return 1;
	}else{
		exit(1);
		//return 0;
	}
}

int main(){
	int n;
	cin >> n;
	syc(n,1,100);
	int a[n];
	int max=0,sum=0;
	
	for(int i=0;i<n;i++){
		cin >> a[i];
		sum+=a[i];
		if(i==0){
			max=a[i];
		}else{
			if(a[i]>max)max=a[i];
		}		
	}

	if((sum-max) > max){
		cout << "Yes" << endl;
	}else{
		cout << "No" << endl;
	}
	return 0;
}