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 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>
using namespace std;

int main() {
   int N;
   int x;
   int maxi_king;
   int max_bs;
   vector <int> sanyiszamok;
   cin >> N;
   for(int i=0;i<N;i++){
     cin >> x;
    sanyiszamok.push_back(x);
   }
   maxi_king=0;
   max_bs=sanyiszamok[0];
   for(int i=1;i<N;i++){
     if(sanyiszamok[i]>max_bs){
       max_bs=sanyiszamok[i];
       maxi_king=i;
     }
   }
   int osszegsanyi=0;
   for(int i=0;i<N;i++){
     osszegsanyi+=sanyiszamok[i];
     
   }  
   osszegsanyi-=max_bs;
   if(max_bs<osszegsanyi){
     cout << "Yes";
   }else{
     cout << "No";
   }
	return 0;
}