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;

const string YES = "Yes";
const string NO = "No";

void solve(int N, std::vector<int> L){
    int temp1=0,temp2=0;
    for(auto it=L.begin();it !=L.end();++it){
        if(temp1<*it){
            temp1=*it;
        }
        temp2+=*it;
        //printf("temp1%d\n",temp1);
        //printf("temp2%d\n",temp2);

    }
    if(temp2>2*temp1){
        printf("Yes\n");
    }else{
        printf("No\n");
    }


}

// Generated by 1.1.4 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
int main(){
    int N;
    scanf("%d",&N);
    std::vector<int> L(N);
    for(int i = 0 ; i < N ; i++){
        scanf("%d",&L[i]);
    }
    solve(N, std::move(L));
    return 0;
}