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<bits/stdc++.h>
using namespace std;
#define arep(i,x,n) for(int i=int(x);i<(int)(n);i++)
#define rep(i,n) for(long long i = 0;i < n;++i)
#define pi 3.141592653589793
#define eps 0.00000001
#define INF 1e9+7  
using ll = long long; 
using P=pair<int,int>;
using lP=pair<ll,ll>;
using fP=pair<double,double>;


int main(){
    int n;
    cin>>n;
    vector<int>l(n);
    rep(i,n)cin>>l[i];
    sort(l.begin(),l.end());
    int sum=0;
    rep(i,n-1)sum+=l[i];
    if(sum>l[n-1])cout<<"Yes"<<endl;
    else cout<<"No"<<endl;
    return 0;
}