#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ld long double
#define show(x,y) cout << y << " " << #x << endl;
#define show2(x,y,i,j) cout << y << " " << #x << " " << j << " " << #i << endl;
#define show3(x,y,i,j,p,q) cout << y << " " << #x << " " << j << " " << #i << " " << q << " " << #p << endl;
#define show4(x,y) for(auto it:x) cout << it << " "; cout << #y << endl;
typedef pair<int,int>pii;
typedef pair<pii,int>pi2;
int n=0;
int len=0;
int pos[205];
int limit[205];
//int memo[205][205][2][205];
//int dp(int l, int r, bool flag, int cur){
//if(cur>200) return 0;
//if(l+1>=r) return 0;
//if(memo[l][r][flag][cur]!=-1) return memo[l][r][flag][cur];
//int ans=0;
//int dist=cur;
//if(!flag) dist+=(pos[l+1]-pos[l]);
//else dist+=(pos[l+1]+k-pos[r]);
//ans=max(ans,dp(l+1,r,0,dist)+(dist<=limit[l+1]));
//int dist2=cur;
//if(flag) dist2+=(pos[r]-pos[r-1]);
//else dist2+=(pos[l]+k-pos[r-1]);
//ans=max(ans,dp(l,r-1,1,dist2)+(dist2<=limit[r-1]));
//return memo[l][r][flag][cur]=ans;
//}
void solve(){
cin >> n >> len;
//memset(memo,-1,sizeof(memo));
for(int x=1;x<=n;x++) cin >> pos[x];
for(int x=1;x<=n;x++) cin >> limit[x];
pos[n+1]=len;
//min amount of time to cover (l,r) such that last is a left/right, visiting k stuff
int dp[n+5][n+5][n+5][2];
for(int x=0;x<n+5;x++){
for(int y=0;y<n+5;y++){
for(int i=0;i<n+5;i++){
dp[x][y][i][0]=1e15;
dp[x][y][i][1]=1e15;
}
}
}
dp[0][n+1][0][0]=0;
dp[0][n+1][0][1]=0;
for(int l=0;l<=n;l++){
for(int r=n+1;r>l;r--){
for(int k=0;k<=n;k++){
//last is left
int dist=dp[l][r][k][0];
//visit lft rgt
int left=pos[l+1]-pos[l];
int right=pos[l]+len-pos[r-1];
//no take
dp[l+1][r][k][0]=min(dp[l+1][r][k][0],dist+left);
dp[l][r-1][k][1]=min(dp[l][r-1][k][1],dist+right);
//take
if(dist+left<=limit[l+1]){
dp[l+1][r][k+1][0]=min(dp[l+1][r][k+1][0],dist+left);
}
if(dist+right<=limit[r-1]){
dp[l][r-1][k+1][1]=min(dp[l][r-1][k+1][1],dist+right);
}
//last is right
dist=dp[l][r][k][1];
left=pos[l+1]+len-pos[r];
right=pos[r]-pos[r-1];
dp[l+1][r][k][0]=min(dp[l+1][r][k][0],dist+left);
dp[l][r-1][k][1]=min(dp[l][r-1][k][1],dist+right);
if(dist+left<=limit[l+1]){
dp[l+1][r][k+1][0]=min(dp[l+1][r][k+1][0],dist+left);
}
if(dist+right<=limit[r-1]){
dp[l][r-1][k+1][1]=min(dp[l][r-1][k+1][1],dist+right);
}
//show3(l,l,r,r,k,k);
//show2(left,dp[l][r][k][0],right,dp[l][r][k][1]);
}
}
}
int best=0;
for(int l=0;l<=n;l++){
for(int r=l+1;r<=n+1;r++){
for(int k=0;k<=n;k++){
if(dp[l][r][k][0]<1e15) best=max(best,k);
if(dp[l][r][k][1]<1e15) best=max(best,k);
}
}
}
cout << best;
}
int32_t main(){
ios::sync_with_stdio(0);
cin.tie(0);
int t=1;
//cin >> t;
while(t--){
solve();
}
}
