Codeforces Round #396 (Div. 2) B

Mahmoud has n line segments, the i-th of them has length ai. Ehab challenged him to use exactly 3 line segments to form a non-degenerate triangle. Mahmoud doesn't accept challenges unless he is sure he can win, so he asked you to tell him if he should accept the challenge. Given the lengths of the line segments, check if he can choose exactly 3 of them to form a non-degenerate triangle.

Mahmoud should use exactly 3 line segments, he can't concatenate two line segments or change any length. A non-degenerate triangle is a triangle with positive area.

Input

The first line contains single integer n (3 ≤ n ≤ 10⁵) — the number of line segments Mahmoud has.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 10⁹) — the lengths of line segments Mahmoud has.

Output

In the only line print "YES" if he can choose exactly three line segments and form a non-degenerate triangle with them, and "NO" otherwise.

Examples

input

5
1 5 3 2 4

output

YES

input

3
4 1 2

output

NO

Note

For the first example, he can use line segments with lengths 2, 4 and 5 to form a non-degenerate triangle.

题意：选出三个数字看能不能构成三角形

解法：

1 根据三角形的特点，排序求出两小点的和以及两大点的差，符合条件就输出

2 好吧，貌似只要找出两小点的和大于第三点就行

#include<bits/stdc++.h>
using namespace std;
int a[200000];
int frsum[200000];
int afsum[200000];
int main(){
    int n;
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>a[i];
    }
    sort(a+1,a+1+n);
    for(int i=2;i<=n-1;i++){
        frsum[i]=a[i-1]+a[i];
        afsum[i]=a[i+1]-a[i];
    }
    for(int i=2;i<=n-1;i++){
        if(frsum[i]>a[i]&&afsum[i]<a[i-1]){
            cout<<"YES"<<endl;
            return 0;
        }
    }
    cout<<"NO"<<endl;
    return 0;
}