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 <iostream>
#include <vector>
#include <boost/range/numeric.hpp>
#include <boost/range/algorithm/max_element.hpp>
#include <boost/range/irange.hpp>
#include <boost/foreach.hpp>
using namespace std;


int main() {
    int n;
    cin >> n;
    vector<int> vec(n);
    for (int i : boost::irange(0, n, 1)) {
        cin >> vec.at(i);
    }

    long sum = boost::accumulate(vec, 0);
    int max_num = *(boost::max_element(vec));
    if (max_num < sum - max_num) {
        cout << "Yes" << endl;
    } else {
        cout << "No" << endl;
    }
    return 0;

}