CodeForces - 426A(排序)

Sereja and Mugs

Time Limit: 1000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

Submit Status

Description

Sereja showed an interesting game to his friends. The game goes like that. Initially, there is a table with an empty cup and n water mugs on it. Then all players take turns to move. During a move, a player takes a non-empty mug of water and pours all water from it into the cup. If the cup overfills, then we assume that this player lost.

As soon as Sereja's friends heard of the game, they wanted to play it. Sereja, on the other hand, wanted to find out whether his friends can play the game in such a way that there are no losers. You are given the volumes of all mugs and the cup. Also, you know that Sereja has (n - 1)friends. Determine if Sereja's friends can play the game so that nobody loses.

Input

The first line contains integers n and s(2 ≤ n ≤ 100; 1 ≤ s ≤ 1000) — the number of mugs and the volume of the cup. The next line contains n integers a₁, a₂, ..., a_n(1 ≤ a_i ≤ 10). Number a_i means the volume of the i-th mug.

Output

In a single line, print "YES" (without the quotes) if his friends can play in the described manner, and "NO" (without the quotes) otherwise.

Sample Input

Input

3 4
1 1 1

Output

YES

Input

3 4
3 1 3

Output

YES

Input

3 4
4 4 4

Output

NO

Source

Codeforces Round #243 (Div. 2)

题意：一个空杯，有n杯水，n-1个人每个人选一杯水往一个瓶子里倒水，不溢出就YES。

题解：扔掉水最多的那一杯，剩下的和如果小于等于瓶子容积就YES。

#include <iostream>
#include <algorithm>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <stdio.h>
using namespace std;
int a[];
int main()
{
    int i,n,t,c,sum=;
    scanf("%d%d",&n,&c);
    for(i=;i<n;i++)
    {
        cin>>a[i];
        sum+=a[i];
    }
    sort(a,a+n);
    sum-=a[n-];
    if(sum<=c)
    cout<<"YES"<<endl;
    else
    cout<<"NO"<<endl;

    return ;
}