//#pragma GCC optimize("O3")
//#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template <class T>
using Tree =
tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define int long long
#define ll long long
#define pii pair<int, int>
#define pll pair<ll, ll>
#define fi first
#define se second
#define rep(i, a, b) for (int i = a; i <= b; ++i)
#define per(i, b, a) for (int i = b; i >= a; --i)
#define pb push_back
#define eb emplace_back
#define all(v) (v).begin(), (v).end()
#define lsb(x) (x)&(-x)
#define bug(x) cerr << #x << " " << x << endl;
void setIO(string name = "") {
ios_base::sync_with_stdio(false);
cin.tie(nullptr); cout.tie(nullptr);
if (!name.empty()) {
freopen((name + ".in").c_str(), "r", stdin);
freopen((name + ".out").c_str(), "w", stdout);
}
}
ll fexp(ll a, ll b, ll m) {
if (b == 0) return 1LL;
ll p = a;
ll ans = 1;
while (b > 0) {
if (b % 2 != 0) ans = (ans * p) % m;
p = (p * p) % m;
b >>= 1;
}
return ans;
}
const int MAXN = 2e2 + 10;
const int INF = 1e18 + 5;
const int MOD = 1e9+7;
int dp[2*MAXN][2*MAXN][MAXN][2], a[2*MAXN], x[MAXN], tt[MAXN], t[2*MAXN];
void solve() {
int N, L;cin >> N >> L;
rep(i,1,N) cin >> x[i];
rep(i,1,N) cin >> tt[i];
int n = 2*N+1;
rep(i,1,N) a[i] = x[i] - L, t[i] = tt[i];
a[N+1] = 0, t[N+1] = -1;
rep(i,N+2,2*N+1) a[i] = x[i-N-1], t[i] = tt[i-N-1];
rep(i,1,n) rep(j,1,n) rep(k,0,n) dp[i][j][k][0] = dp[i][j][k][1] = INF;
dp[N+1][N+1][0][0] = dp[N+1][N+1][0][1] = 0;
int ans = 0;
rep(sz,1,N) {
rep(l,1,n-sz) {
rep(k,0,n) {
int r = l + sz;
int t1 = a[l+1]-a[l], t2 = a[r]-a[l];
int t3 = a[r] - a[r-1];
dp[l][r][k][0] = min({dp[l+1][r][k][0] + t1, dp[l+1][r][k][1] + t2, dp[l][r][k][0]});
if (k>0 && dp[l+1][r][k-1][0] + t1 <= t[l]) dp[l][r][k][0] = min(dp[l][r][k][0], dp[l+1][r][k-1][0] + t1);
if (k > 0 && dp[l+1][r][k-1][1] + t2 <= t[l]) dp[l][r][k][0] = min(dp[l][r][k][0], dp[l+1][r][k-1][1] + t2);
dp[l][r][k][1] = min({dp[l][r-1][k][1] + t3, dp[l][r-1][k][0] + t2, dp[l][r][k][1]});
if (k>0 && dp[l][r-1][k-1][1] + t3 <= t[r]) dp[l][r][k][1] = min(dp[l][r][k][1], dp[l][r-1][k-1][1] + t3);
if (k>0 && dp[l][r-1][k-1][0] + t2 <= t[r]) dp[l][r][k][1] = min(dp[l][r][k][1], dp[l][r-1][k-1][0] + t2);
}
}
}
rep(i,1,N) {
rep(k,1,n) {
if (min(dp[i][i+N][k][0] , dp[i][i+N][k][1]) != INF) {
if (ans < k) {
ans = k;
}
}
}
}
cout << ans << '\n';
}
int32_t main() {
setIO();
int tt = 1;
//cin >> tt;
while (tt-- > 0) solve();
return 0;
}
