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 <stdio.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <numeric>
#include <string.h>
using namespace std;
typedef long long int lli;
#define urept(soeji, start, n) for (int soeji = start; soeji < n; soeji++)
#define drept(soeji, start, n) for (int soeji = start; soeji > n; soeji--)
int main(void)
{
    int N;
    cin >> N;
    int *L;
    L = new int[N];
    urept(i, 0, N)
    {
        cin >> L[i];
    }
    sort(L, L + N);
    int sum = 0;
    urept(i, 0, N - 1)
    {
        sum += L[i];
    }
    if (sum > L[N - 1])
    {
        cout << "Yes" << endl;
    }
    else
    {
        cout << "No" << endl;
    }
    return 0;
}