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: Wrong Answer)

#include <cstdio>
#include <iostream>
#include <vector>
#include <set>
#include <cstring>
#include <string>
#include <map>
#include <cmath>
#include <ctime>
#include <algorithm>
#include <bitset>
#include <queue>
#include <sstream>
using namespace std;
#define mp make_pair
#define pb push_back
#define re return
#define fi first
#define se second
typedef vector<int> vi;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef vector<pii> vpii;
typedef vector<string> vs;
typedef vector<vi> vvi;
int n,a[11];
int main() {
    //freopen("input.txt","r",stdin);
	//freopen("output.txt","w",stdout);
	cin>>n;
	int sum=0;
	int mx=0;
	for(int i=0;i<n;i++){
		cin>>a[i];
		sum+=a[i];
		mx=max(mx,a[i]);
	}sum-=mx;
	if(mx>sum)cout<<"Yes"<<endl;
	else cout<<"No"<<endl;
    return 0;
}