codeforces 766 B

B. Mahmoud and a Triangle

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

问题就是给出边长，问是否能构成三角形。

两种方法：第一种就是排序啊a
+a[n+1]>a[n+2]就输出YES。（nlogn）
第二种：可以观察到斐波那契数列是符合题意的最长数列，而到fib（50）时已经超过题目范围的边长10^9。因此当n>50时一定有解。n<50时则暴力n^3或采用第一种方法算出解。
这里的代码是第一种的方法。

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=100000;
int n,a[maxn+10];
int main(){
scanf("%d",&n);
for(int i = 1; i <= n; i++){
scanf("%d",&a[i]);
}
sort(a+1,a+n+1);
for(int i = 1; i <= n-2; i++){
if(a[i] + a[i+1] > a[i+2]){
printf("YES");
return 0;
}
}
printf("NO");
return 0;
}