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)

// header {{{
#include <bits/stdc++.h>
using namespace std;

// {U}{INT,LONG,LLONG}_{MAX,MIN}
#define INF         INT_MAX/3
#define LLINF       LLONG_MAX/3
#define MOD         (1000000007LL)
#define MODA(a, b)  a=((a)+(b))%MOD
#define MODP(a, b)  a=((a)*(b))%MOD
#define inc(i, l, r)   for(int i=(l);i<(r);i++)
#define dec(i, l, r)   for(int i=(r)-1;i>=(l);i--)
#define pb          push_back
#define se          second
#define fi          first
#define mset(a, b)  memset(a, b, sizeof(a))

using LL  = long long;
using G   = vector<vector<int>>;

int di[] = {0, -1, 0, 1};
int dj[] = {1, 0, -1, 0};
// }}}

int main() {
    cin.tie(0);ios::sync_with_stdio(false);
    int n;cin >> n;
    int mmax = 0;
    int sum = 0;
    inc(i, 0, n){
        int l;cin >> l;
        mmax = max(mmax, l);
        sum += l;
    }
    cout << (mmax < sum-mmax ? "Yes" : "No") << endl;
    return 0;
}