#include <iostream>
#include <vector>
using namespace std;
const int N = 200;
const long long INF = 1e18;
long long dp[1+N][1+N][2][1+N];
// dp[l][r][p][s] is minimum time where:
// - [l, r] is range of stamps left
// - p is the current position (boolean - either l-1 or r+1)
// - s is the current number of stamps collected
int main() {
int n, L;
cin >> n >> L;
vector<int> x(1+n), t(1+n);
for (int i = 1; i <= n; ++i) cin >> x[i];
for (int i = 1; i <= n; ++i) cin >> t[i];
x[0] = 0, t[0] = -1; // add stamp at x=0 that cannot be collected
auto d = [&x, &n, &L](int i, int j) {
if (i > j) swap(i, j);
return min(x[j]-x[i], x[i]+L-x[j]);
};
// fill with INF
for (int l = 0; l <= n; ++l) {
for (int r = max(0, l-1); r <= n; ++r) {
for (int p = 0; p <= 1; ++p) {
for (int s = 0; s <= n; ++s) dp[l][r][p][s] = INF;
}
}
}
// addition of extra stamp makes base case easier
dp[1][n][0][0] = dp[1][n][1][0] = 0;
/* answers will be stored at sz = 0 (r = l-1), but we do not traverse
since they are final states - they do not need to contribute to other states */
// dp - pushing answers to later states rather than pulling from previous
for (int sz = n; sz >= 1; --sz) {
for (int l = 1; l+sz-1 <= n; ++l) {
int r = l+sz-1, before = (l-1+(n+1))%(n+1), after = (r+1)%(n+1);
for (int s = 0; s <= n; ++s) {
// p = 0 means at l-1, p = 1 means at r+1
for (int p = 0; p <= 1; ++p) { // current position
for (int p1 = 0; p1 <= 1; ++p1) { // next position
int cur = (!p ? before : after); // current location
int nxt = (!p1 ? l : r); // next location
long long time = dp[l][r][p][s] + d(cur, nxt); // total time taken
bool delta = time <= t[nxt]; // extra stamps collected (0 or 1)
int l1 = l + (p1 == 0), r1 = r - (p1 == 1), s1 = s+delta;
if (l1 > n) continue; // r1 < 0 will never happen since l >= 1
dp[l1][r1][p1][s1] = min(dp[l1][r1][p1][s1], time);
}
}
}
}
}
int maxCollected = 0;
for (int s = 1; s <= n; ++s) {
for (int l = 1; l <= n; ++l) {
if (dp[l][l-1][0][s] != INF || dp[l][l-1][1][s] != INF) {
maxCollected = s;
}
}
}
cout << maxCollected << endl;
return 0;
}
