codeforces

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 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

题意：给定n个数，问能否找到三个数可以构成三角形

题解：从大到小判断相邻三个即可

代码：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <bitset>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>
#include <list>
#include <set>
#include <map>
#define rep(i,a,b) for(int i = a;i <= b;++ i)
#define per(i,a,b) for(int i = a;i >= b;-- i)
#define mem(a,b) memset((a),(b),sizeof((a)))
#define FIN freopen("in.txt","r",stdin)
#define FOUT freopen("out.txt","w",stdout)
#define IO ios_base::sync_with_stdio(0),cin.tie(0)
#define mid ((l+r)>>1)
#define ls (id<<1)
#define rs ((id<<1)|1)
#define N 100005
#define INF 0x3f3f3f3f
#define INFF ((1LL<<62)-1)
using namespace std;
typedef long long LL;
typedef pair<int, int> PIR;
const double eps = 1e-8;

int n, a[N];
int main()
{IO;
    while(cin >> n){
        rep(i, 1, n)    cin >> a[i];
        sort(a+1, a+n+1);
        int ok = 0;
        per(i, n, 3){
            int x = a[i], y = a[i-1], z = a[i-2];
            if(x+y > z && x+z > y && y+z > x){
                ok = 1;
                break;
            }
        }
        cout << (ok ? "YES" : "NO") << endl;
    }
    return 0;
}