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 <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <map>
#include <utility>
#define endl '\n'
#define fi first
#define se second
#define pb push_back
#define debug(x) cout << "[debug] " << #x << ": " << x << endl
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
const int MOD = 1e9 + 7;
const int maxn = 1e6 + 5;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-8;

int main() {
    int n; cin >> n;
    vector<int> nums;
    for(int i = 0; i < n; ++i){
    	int x; cin >> x;
    	nums.pb(x);
    }
    sort(nums.begin(), nums.end());
    int y = nums[n - 1];
    int sum = 0;
    for(int i = 0; i < n - 1; ++i) sum += nums[i];
    if(sum > y) cout << "Yes" << endl;
	else cout << "No" << endl;
    return 0;
}