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最新推荐文章于 2025-05-22 14:55:36 发布

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最新推荐文章于 2025-05-22 14:55:36 发布

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分类专栏： ACM

本文链接：https://blog.youkuaiyun.com/wu_yihao/article/details/18890249

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George decided to prepare a Codesecrof round, so he has prepared m problems for the round. Let's number the problems with integers 1 through m. George estimates the i-th problem's complexity by integer b_i.

To make the round good, he needs to put at least n problems there. Besides, he needs to have at least one problem with complexity exactly a₁, at least one with complexity exactly a₂, ..., and at least one with complexity exactly a_n. Of course, the round can also have problems with other complexities.

George has a poor imagination. It's easier for him to make some already prepared problem simpler than to come up with a new one and prepare it. George is magnificent at simplifying problems. He can simplify any already prepared problem with complexity c to any positive integer complexity d (c ≥ d), by changing limits on the input data.

However, nothing is so simple. George understood that even if he simplifies some problems, he can run out of problems for a good round. That's why he decided to find out the minimum number of problems he needs to come up with in addition to the m he's prepared in order to make a good round. Note that George can come up with a new problem of any complexity.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 3000) — the minimal number of problems in a good round and the number of problems George's prepared. The second line contains space-separated integers a₁, a₂, ..., a_n (1 ≤ a₁ < a₂ < ... < a_n ≤ 10⁶) — the requirements for the complexity of the problems in a good round. The third line contains space-separated integers b₁, b₂, ..., b_m (1 ≤ b₁ ≤ b₂... ≤ b_m ≤ 10⁶) — the complexities of the problems prepared by George.

Output

Print a single integer — the answer to the problem.

Sample test(s)

Input

3 5
1 2 3
1 2 2 3 3

Output

Input

3 5
1 2 3
1 1 1 1 1

Output

Input

3 1
2 3 4
1

Output

Note

In the first sample the set of the prepared problems meets the requirements for a good round.

In the second sample, it is enough to come up with and prepare two problems with complexities 2 and 3 to get a good round.

In the third sample it is very easy to get a good round if come up with and prepare extra problems with complexities: 2, 3, 4.

双指针操作

#include <cstdio>

int a[3010];
int b[3010];

int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++)
		scanf("%d",a+i);
	for (int i=1;i<=m;i++)
		scanf("%d",b+i);

	int i=1,j=1;
	int ans = 0;
	while (i<=n&&j<=m)
	{
		if (a[i]<=b[j])
			i++,j++,ans++;
		else
			j++;
	}
	int aans = n-ans;
	if (aans < 0)
		aans = 0;
	printf("%d",aans);
}