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 PI 3.141592653589793238462643383279
typedef long long ll;
const long double EPS = 1e-9;
const ll MOD = 1e9+7;
#define N 111
int l[N];

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout << setprecision(20);
    #ifdef LOCAL_TEST
    freopen("input.txt","r",stdin);
    #endif
    int n; cin >> n;
    int s = 0;
    for(int i=0;i<n;i++) cin >> l[i], s += l[i];
    sort(l, l + n);
    cout << ((2 * l[n - 1] < s) ? "Yes" : "No") << '\n';
    return 0;
}