Codeforces 839B-Game of the Rows

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, the i-th group contains ai 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 nrows, each of them has 8 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 and k (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, ..., ak (1 ≤ ai ≤ 10000), where ai denotes the number of soldiers in the i-th group.

It is guaranteed that a1 + a2 + ... + ak ≤ 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).

题意：将n批人安排进题目里给出的那种n行座位中，不同批的人只能隔着坐，问能不能安排下这些人

解题思路：首先必然先安排大于等于3人的批次，将他们尽量安排在中间四连座的座位，安排不下了再安排在旁边两连座的，若也安排不下了，则必然输出NO，然后在安排剩余2个人的批次，先尽量安排在两连座的，若安排不下了则安排在中间四连座的，并且还可以安排边上坐一个人，若也安排不下，则将两个人拆开当作一个人去安排，最后判断座位够不够剩余1个人的批次即可

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int n, k, x[10009];int have[5], sum[5];int main(){while (~scanf("%d%d", &n, &k)){int flag = 1;memset(sum, 0, sizeof sum);memset(have, 0, sizeof have);have[2] = 2 * n;have[4] = n;for (int i = 0; i < k; i++) scanf("%d", &x[i]);for (int i = 0; i < k; i++){while (x[i] >= 3){if (have[4]){have[4]--;x[i] -= 4;}else if (have[2]){have[2]--;x[i] -= 2;}else { flag = 0; printf("NO\n"); break; }}sum[x[i]]++;}while (sum[2]){if (have[2]){have[2]--;sum[2]--;}else if (have[4]){have[4]--;sum[2]--;have[1]++;}else{sum[2]--;sum[1] += 2;}}if (flag){if (sum[1] > have[1] + have[2] + have[4] * 2) printf("NO\n");else printf("YES\n");}}return 0;}