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.
# https://bigocoder.com/courses/74/lectures/1139/problems/522?view=statement
# O(n + m)
def practice2():
    n, m = map(int, input().split())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))
    cnt = index = 0

    # 2 pointers
    while cnt < n and index < m:
        # if component in a < b', move on
        if a[cnt] <= b[index]:
            cnt += 1
        # if not? ==> a > b, increase b to compare a <> b[i+1]
        index += 1

    # total - (excluded records) = result records
    print(n - cnt)


if __name__ == "__main__":
    practice2()
