/*
Keep track of prefix maximum and minimum as these things tell us the highest and the lowest points we reached so far with the snowballs
For each pair of two adjacent snowballs, we should binary search the point x at which maxq[x] + abs(minq[x]) >= v[i+1] - v[i] and then add accordingly
to each end based on where the last update came from
We should deal separately with the two extremes too
*/
#include <bits/stdc++.h>
using namespace std;
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n, q;
cin >> n >> q;
vector<long long> v(n+1);
for(int i = 1; i <= n; i++)
cin >> v[i];
vector<long long> minq(q+1), maxq(q+1);
long long sum = 0;
for(int i = 1; i <= q; i++)
{
long long x;
cin >> x;
sum += x;
minq[i] = min(minq[i-1], sum);
maxq[i] = max(maxq[i-1], sum);
}
vector<long long> ans(n+1);
for(int i = 1; i < n; i++)
{
int st = 1;
int dr = q;
int poz = 0;
while(st <= dr)
{
int mid = (st + dr) / 2;
if(abs(minq[mid]) + abs(maxq[mid]) < v[i+1] - v[i])
{
poz = mid;
st = mid + 1;
}
else
dr = mid - 1;
}
ans[i] += maxq[poz];
ans[i+1] += abs(minq[poz]);
if(poz != q && maxq[poz+1] > maxq[poz])
ans[i] += (v[i+1] - v[i]) - (abs(minq[poz]) + abs(maxq[poz]));
if(poz != q && minq[poz+1] < minq[poz])
ans[i+1] += (v[i+1] - v[i]) - (abs(minq[poz]) + abs(maxq[poz]));
}
ans[1] += abs(minq[q]);
ans[n] += abs(maxq[q]);
for(int i = 1; i <= n; i++)
cout << ans[i] << '\n';
}
