Codeforces Round #396(Div. 2)B. Mahmoud and a Triangle【上古时代题】nlogn判断三角形

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本文介绍了一个算法问题，即如何从给定的线段中找出能否构成一个非退化的三角形的方法。通过将线段按长度排序，并检查每三个连续线段是否满足构成三角形的条件，来快速解决这个问题。

Mahmoud has n line segments, the i-th of them has length a_i. 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, ..., a_n (1 ≤ a_i ≤ 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条边，让你判断是否能够用其中三条边整出来一个三角形。

思路：

直接将边按照从小到大排序。然后相邻的三条边进行判断即可。

大致理由：

此时有三条相邻的边：a，b，c。显然排序之后是a<b<c的，如果此时a+b<c，那么肯定a+b<d.e.f.g.h..................，肯定a和b不能与c之后的边组成三角形。

那么同理，可以递推出其他情况。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[100060];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)scanf("%d",&a[i]);
        sort(a+1,a+1+n);
        int flag=0;
        for(int i=1;i<=n;i++)
        {
            if(i+1<=n&&i+2<=n)
            if(a[i]+a[i+1]>a[i+2])
            {
                flag=1;
            }
        }
        if(flag==1)printf("YES\n");
        else printf("NO\n");
    }
}