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 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <algorithm>
#include <utility>
#include <iomanip>
#include <functional>
#include <queue>
#include <stack>
#include <map>
#include <vector>
#include <cmath>
#include <cctype>
#include <cstdio>
#define ll long long
#define INF 1e+18
#define MOD 1000000007
using namespace std;



int main(){
    int N;
    cin >> N;
    int L[N];
    for(int i=0; i<N; i++){
        cin >> L[i];
    }
    sort(L,L+N);
    for(int i=0; i<N-1; i++){
        L[N-1]-=L[i];
    }
    if(L[N-1]>=0){
        cout<<"No"<<endl;
    }else{
        cout<<"Yes"<<endl;
    }
    return 0;
}