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*<sub class="lower-index">*i*</sub>.

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 *a*<sub class="lower-index">1</sub>, at least one with complexity exactly *a*<sub class="lower-index">2</sub>, ..., and at least one with complexity exactly *a*<sub class="lower-index">*n*</sub>. 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.
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 *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub> (1 ≤ *a*<sub class="lower-index">1</sub> &lt; *a*<sub class="lower-index">2</sub> &lt; ... &lt; *a*<sub class="lower-index">*n*</sub> ≤ 10<sup class="upper-index">6</sup>) — the requirements for the complexity of the problems in a good round. The third line contains space-separated integers *b*<sub class="lower-index">1</sub>, *b*<sub class="lower-index">2</sub>, ..., *b*<sub class="lower-index">*m*</sub> (1 ≤ *b*<sub class="lower-index">1</sub> ≤ *b*<sub class="lower-index">2</sub>... ≤ *b*<sub class="lower-index">*m*</sub> ≤ 10<sup class="upper-index">6</sup>) — the complexities of the problems prepared by George.
# 387B >> accepted
# chu giai giai thuat: vi 2 list n va m da duoc sap xep tang dan, cho i chay theo n, con j chay theo m.
# khi j chay tu 0>>m, neu required problem i co do phuc tap n_list[i] nho hon problem j co do phuc tap m_list[j] thi
# thoa dieu kien de bai, tuc la m[j] co the giam xuong n[i], ta tang bien dem, dong thoi tang i de xet cap i,j tiep theo

ip = input()
n = int(ip.split()[0]) # number of required problems
m = int(ip.split()[1]) # number of prepared problems
n_list = list(map(int,input().split()))
# n_list.sort() # tai sao ko .sort() o ben tren duoc luon.
# n_list.reverse()
m_list = list(map(int,input().split()))
# m_list.sort()
# m_list.reverse()
count = 0
i = 0
# Truong hop: m>=n
for j in range(m):
    if n_list[i]<=m_list[j]:
        count+=1
        i+=1
#         print ('count',count)
#         print('i',i)
#         print ('j',j)
        # dieu kien dung vong lap khi m<n
    if i>=(n):
        break
print (max(0,n-count))