//自己想到的，好開心(對照過提解了，確定是對的)
#include <bits/stdc++.h>
using namespace std;
#define ll long long
int n, l, a[210], t[210];
ll inf = 1e13;
ll dp[210][210][210][2]; //dp[l][r][k][w]代表逆時針走過l，順時針走過r，共拿過k個郵票，現在在哪個端點(w=0是左)，所需的最短時間
int walk (int st, int ed) {
if (st == ed) return 0;
if (st > ed) swap(st, ed);
return min((a[st] + l - a[ed]), a[ed] - a[st]);
}
signed main () {
ios_base::sync_with_stdio(false); cin.tie(0);
memset(dp, 63, sizeof(dp));
dp[0][0][0][0] = dp[0][0][0][1] = 0;
cin >> n >> l;
for (int i = 1; i <= n; i++) cin >> a[i];
for (int i = 1; i <= n; i++) cin >> t[i];
a[0] = 0, a[n + 1] = l, t[0] = t[n + 1] = -1;
int ans = 0;
for (int l = 0; l <= n; l++) {
for (int r = 0; r <= n and l + r <= n; r++) {
for (int k = 0; k <= n; k++) {
if (min(dp[l][r][k][1], dp[l][r][k][0]) < inf) {
// if (k > ans) cout << l << ' ' << r << ' ' << k << ' ' << dp[l][r][k][0] << ' ' << dp[l][r][k][1] << '\n';
ans = max(ans, k);
}
//逆時針
if (l + 1 <= n){
ll tt = dp[l][r][k][0] + walk(n + 1 - l, n + 1 - (l + 1));
tt = min(tt, dp[l][r][k][1] + walk(r, n + 1 - (l + 1)) );
if (tt <= t[n + 1 - (l + 1)]) dp[l + 1][r][k + 1][0] = min(dp[l + 1][r][k + 1][0], tt);
else dp[l + 1][r][k][0] = min(dp[l + 1][r][k][0], tt);
}
//順時針
if (r + 1 <= n){
ll tt = dp[l][r][k][0] + walk(n + 1 - l, r + 1);
tt = min(tt, dp[l][r][k][1] + walk(r, r + 1) );
if (tt <= t[r + 1]) dp[l][r + 1][k + 1][1] = min(dp[l][r + 1][k + 1][1], tt);
else dp[l][r + 1][k][1] = min(dp[l][r + 1][k][1], tt);
}
}
}
}
cout << ans << '\n';
return 0;}
