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 Rust (1.42.0) (judge verdict: Wrong Answer)

use std::io::{stdin, Read};
use std::str::FromStr;
use std::cmp::Reverse;

fn read<T: FromStr>() -> Option<T> {
    let stdin = stdin();
    let s = stdin
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect::<String>();
    s.parse::<T>().ok()
}

fn main() {
    let n: i64 = read().unwrap();
    let mut ls: Vec<i64> = Vec::new();
    for _i in 0..n {
        let l: i64 = read().unwrap();
        ls.push(l);
    }
    ls.sort_by_key(|&x| Reverse(x));
    let maxl: i64 = ls[0];
    let suml = ls.iter().skip(1).fold(0, |acc, i| acc + i);
    if maxl < suml {
        println!("YES");
    } else {
        println!("NO");
    }
}