codeforces387B

George and Round

 CodeForces - 387B 

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 mhe'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.

Examples

Input

3 5
1 2 3
1 2 2 3 3

Output

0

Input

3 5
1 2 3
1 1 1 1 1

Output

2

Input

3 1
2 3 4
1

Output

3

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.

 

sol：注意到n很小，n²都可以过，于是直接暴力模拟。

我猜应该有O(n*logn)的做法，比方说开一个multiset，先排遍序，每次取大于等于当前这个数的第一个，然后弹掉（这只是嘴巴，我没写过）

#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=0;
    bool f=0;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0)
    {
        putchar('-'); x=-x;
    }
    if(x<10)
    {
        putchar(x+'0');    return;
    }
    write(x/10);
    putchar((x%10)+'0');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=3005;
int n,m,a[N],b[N];
bool Bo[N];
int main()
{
    int i,j,ans;
    R(n); R(m);
    ans=n;
    for(i=1;i<=n;i++) R(a[i]);
    for(i=1;i<=m;i++) R(b[i]);
    sort(b+1,b+m+1);
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=m;j++) if(b[j]>=a[i]&&(!Bo[j]))
        {
            Bo[j]=1; ans--; break;
        }
    }
    Wl(ans);
    return 0;
}