#include <bits/stdc++.h>
using namespace std;

int n;
vector<long long> a, b;  // a[0..n-1], b[0..n-1] da sap xep khong giam

// Nhan xet: chen so w = 0 luon tot nhat (mang luon nho hon hoac bang moi cach chen khac).
// Sau k lan chen 0, mang cuoi cung la: k so 0, roi a[0..n-k-1]. Can c[i] <= b[i] voi moi i.
// Dieu kien tang theo k nen co the tim k nho nhat.

// Sub 1: mo phong truc tiep. Thu k = 0, 1, 2, ...: thuc hien k lan chen 0 (them 0, sap xep, bo phan tu lon nhat)
// roi kiem tra a[i] <= b[i]. Dung khi n <= 100.
void sub1() {
    for (int k = 0;; k++) {
        vector<long long> c = a;
        for (int t = 0; t < k; t++) {
            c.push_back(0);
            sort(c.begin(), c.end());
            c.pop_back();  // bo phan tu lon nhat
        }
        bool ok = true;
        for (int i = 0; i < n && ok; i++)
            if (c[i] > b[i]) ok = false;
        if (ok) {
            cout << k << "\n";
            return;
        }
    }
}

// Sub 2: nhi phan theo k. Voi k lan chen 0, mang cuoi la 0 (k phan tu dau) va a[0..n-k-1] tiep theo,
// nen can a[j] <= b[j+k] voi moi j = 0..n-k-1. Dieu kien don dieu theo k, tim k nho nhat bang tim kiem nhi phan. O(n log n).
bool okK(int k) {
    for (int j = 0; j + k < n; j++)
        if (a[j] > b[j + k]) return false;
    return true;
}
void sub2() {
    int lo = 0, hi = n;  // k = n luon dung (mang toan 0)
    while (lo < hi) {
        int mid = (lo + hi) / 2;
        if (okK(mid)) hi = mid;
        else lo = mid + 1;
    }
    cout << lo << "\n";
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> n;
    a.resize(n);
    b.resize(n);
    for (auto &x : a) cin >> x;
    for (auto &x : b) cin >> x;
    if (n <= 100) sub1();  // n nho: mo phong truc tiep
    else sub2();           // n lon: nhi phan
    return 0;
}