Codeforces Round #428 (Div. 2) B. Game of the Rows （思维）

B. Game of the Rows

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Daenerys Targaryen has an army consisting of k groups of soldiers, thei-th group contains a_i soldiers. She wants to bring her army to the other side of the sea to get the Iron Throne. She has recently bought an airplane to carry her army through the sea. The airplane has n rows, each of them has8 seats. We call two seats neighbor, if they are in the same row and in seats{1, 2}, {3, 4}, {4, 5}, {5, 6} or {7, 8}.

A row in the airplane

Daenerys Targaryen wants to place her army in the plane so that there are no two soldiers from different groups sitting on neighboring seats.

Your task is to determine if there is a possible arranging of her army in the airplane such that the condition above is satisfied.

Input

The first line contains two integers n andk (1 ≤ n ≤ 10000,1 ≤ k ≤ 100) — the number of rows and the number of groups of soldiers, respectively.

The second line contains k integers a₁, a₂, a₃, ..., a_k (1 ≤ a_i ≤ 10000), wherea_i denotes the number of soldiers in thei-th group.

It is guaranteed that a₁ + a₂ + ... + a_k ≤ 8·n.

Output

If we can place the soldiers in the airplane print "YES" (without quotes). Otherwise print "NO" (without quotes).

You can choose the case (lower or upper) for each letter arbitrary.

Examples

Input

2 25 8

Output

YES

Input

1 27 1

Output

NO

Input

1 24 4

Output

YES

Input

1 42 2 1 2

Output

YES

Note

In the first sample, Daenerys can place the soldiers like in the figure below:

In the second sample, there is no way to place the soldiers in the plane since the second group soldier will always have a seat neighboring to someone from the first group.

In the third example Daenerys can place the first group on seats (1, 2, 7, 8), and the second group an all the remaining seats.

In the fourth example she can place the first two groups on seats (1, 2) and (7, 8), the third group on seats (3), and the fourth group on seats (5, 6).

思路 ： 一步步走，把每一队的人数拆解为4，2，1，记录总数这样1 肯定占两个坐，所以总人数 sum + 1的个数 >= 8*n 就坐不开了，以下就是坐开的情况，如果四个人的数目和一个数目和能把四个的全部坐满，没问题。否则让所有的两个人入座，如果能坐满下，没问题，否则就得拆分两个人了。

       

#include <queue>#include <map>#include <vector>#include <cmath>#include <stack>#include <climits>#include <string>#include <iostream>#include <cstring>#include <algorithm>using namespace std;int main(){    int n,k;    scanf("%d%d",&n,&k);    int x4 = 0,x2 = 0,x1 = 0;    int sum = 0;    for(int i = 0; i < k; i++)    {        int x;        scanf("%d",&x);        sum += x;        x4 += x/4;        x %= 4;        x2 += x/2;        x1 += x%2;    }    if(x1 + sum > n*8)    {        printf("NO\n");        return 0;    }    if(x4 + x1>= n)    {        printf("YES\n");        return 0;    }    int tmp = n - x4 - x1; // 剩余四人座个数    x2 -= n-x4+n*2; // 两个人的入座。    if(x2 <= 0)    {        printf("YES\n");        return 0;    }    x2 *= 2;    if(x2 <= tmp) printf("YES\n");    else printf("NO\n");    return 0;}