#include <bits/stdc++.h>
using namespace std;
#define int long long
int inf = 1e18;
int n,l,x[205],t[205];
int dp[205][205][205][2];
int sp[205];
int dist(int px,int py)
{
return x[py] - x[px];
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> n >> l;
for (int i = 1; i <= n; i++)
cin >> x[i];
x[n + 1] = l;
for (int i = 1; i <= n; i++)
cin >> t[i];
for (int i = 0; i <= n; i++)
for (int j = 0; j <= n; j++)
for (int q = 0; q <= n; q++)
for (int c = 0; c < 2; c++)
dp[i][j][q][c] = inf;
dp[0][0][0][0] = dp[0][0][0][1] = 0;
for (int i = 0; i <= n - 1; i++)
{
for (int j = 0; j <= n - 1 - i; j++)
{
for (int k = 0; k <= i + j; k++)
{
//cout << i << ' ' << j << ' ' << k << ' ' << dp[i][j][k][0] << ' ' << dp[i][j][k][1] << endl;
//cout << i << ' ' << j << ' ' << k << ' ';
int t0 = dp[i][j][k][0] + dist(n - i,n - i + 1);
if (t0 <= t[n - i])
dp[i + 1][j][k + 1][0] = min(dp[i + 1][j][k + 1][0],t0);
else
dp[i + 1][j][k][0] = min(dp[i + 1][j][k][0],t0);
int t1 = dp[i][j][k][0] + l - dist(j + 1,n - i + 1);
if (t1 <= t[j + 1])
dp[i][j + 1][k + 1][1] = min(dp[i][j + 1][k + 1][1],t1);
else
dp[i][j + 1][k][1] = min(dp[i][j + 1][k][1],t1);
int t2 = dp[i][j][k][1] + l - dist(j,n - i);
if (t2 <= t[n - i])
dp[i + 1][j][k + 1][0] = min(dp[i + 1][j][k + 1][0],t2);
else
dp[i + 1][j][k][0] = min(dp[i + 1][j][k][0],t2);
int t3 = dp[i][j][k][1] + dist(j,j + 1);
if (t3 <= t[j + 1])
dp[i][j + 1][k + 1][1] = min(dp[i][j + 1][k + 1][1],t3);
else
dp[i][j + 1][k][1] = min(dp[i][j + 1][k][1],t3);
//cout << t0 << ' ' << t1 << ' ' << t2 << ' ' << t3 << endl;
}
}
}
for (int i = n; i >= 1; i--)
{
bool pot = false;
for (int j = 0; j <= n; j++)
{
int q = n - j;
if (dp[j][q][i][0] < inf or dp[j][q][i][1] < inf)
pot = true;
}
if (pot == true)
{
cout << i;
return 0;
}
}
cout << 0;
return 0;
}
/**
presupun ca exista punctul 0, apoi in dreapta punctele sunt 1,2,...,n iar in stanga sunt n,n-1,...,1
**/
/**
dp[i][j][k][0/1] -> care este timpul minim la care sa fiu la i in stanga (0) sau j in dreapta (1) a.i am acoperit interv [i j] si am k stamp-uri
{i,j,k,0} -> {i + 1,j,k sau k + 1 in functie de timp,0} sau {i,j + 1,k sau k + 1 in functie de timp,1}
{i,j,k,1} -> {i + 1,j,k sau k + 1 in functie de timp,0} sau {i,j + 1,k sau k + 1 in functie de timp,1}
**/
