codeforces 766 B Mahmoud and a Triangle

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题目链接：http://codeforces.com/contest/766/problem/B

题目：

Mahmoud has n line segments, the i-th of them has length ai. Ehab challenged him to useexactly 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.

题目大意：

给你n跟小木棍，问是否能够找到三根使之组成一个三角形

题目分析：

对小木棍长度进行排序，然后暴力扫描就好，注意这里如果2，3，5不能组成三角形，那么就不用再找2，3，+比5大的边了

AC代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define LL long long
const LL maxn=1e5+5;
bool Triangle(LL a,LL b,LL c)
{
    if(a+b>c&&a+c>b&&b+c>a)
        return true;
    return false;
}
LL v[maxn],n;
int main()
{
    while(~scanf("%I64d",&n))
    {
        for(LL i=0;i<n;i++)
            scanf("%I64d",&v[i]);
        sort(v,v+n);
        bool sign=false;
        for(LL i=0;i<n-2;i++)
        {
            if(sign)
                break;
            for(LL j=i+1;j<n-1;j++)
            {
                if(Triangle(v[i],v[j],v[j+1]))
                {
                    sign=true;
                    cout<<"YES"<<endl;
                    break;
                }
                else
                    break;
            }
        }
        if(!sign)
            cout<<"NO"<<endl;
    }
    return 0;
}