#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int MAXN = 2000005; // ← bump this up by a few
ll n, q, a[MAXN], ans[MAXN];
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
cin >> n >> q;
for(int i = 1; i <= n; i++){
cin >> a[i];
}
// 1) Build the prefix‐sum‐of‐nonzeros array
vector<ll> megu;
megu.reserve(q);
while(q--){
ll x;
cin >> x;
if(x == 0) continue;
if(megu.empty()) megu.push_back(x);
else megu.push_back(megu.back() + x);
}
// 2) Extract only the “record” maxima (positive) and minima (negative)
vector<ll> aquaa;
aquaa.reserve(megu.size());
ll bestPos = 0, bestNeg = 0;
for(ll s : megu){
if(s > 0 && s > bestPos){
aquaa.push_back(s);
bestPos = s;
}
else if(s < 0 && llabs(s) > bestNeg){
aquaa.push_back(s);
bestNeg = llabs(s);
}
}
// If there was never a sign‐change, we only have one “record”:
if(aquaa.size() <= 1){
// the only contributions are from the last positive record (bestPos)
// and the last negative record (bestNeg), which go to ans[n] and ans[1].
ans[1] = bestNeg;
ans[n] = bestPos;
for(int i = 1; i <= n; i++){
cout << ans[i] << "\n";
}
return 0;
}
// 3) Keep only the points at which the prefix‐sum flips sign
vector<ll> aqua;
aqua.reserve(aquaa.size());
for(int i = 0; i + 1 < (int)aquaa.size(); i++){
if ((aquaa[i] > 0) != (aquaa[i+1] > 0))
aqua.push_back(aquaa[i]);
}
// always add the final “record”
aqua.push_back(aquaa.back());
// 4) Precompute the dm[] = |aqua[i]| + |aqua[i+1]| array and sort it
vector<ll> dm;
dm.reserve(aqua.size()-1);
for(int i = 0; i + 1 < (int)aqua.size(); i++){
dm.push_back(llabs(aqua[i]) + llabs(aqua[i+1]));
}
sort(dm.begin(), dm.end());
// 5) Sweep through the original a[1..n], splitting each gap a[i+1]-a[i]
// optimally between the two sides
for(int i = 1; i < n; i++){
ll gap = a[i+1] - a[i];
// find the largest dm[p] <= gap
int p = int(upper_bound(dm.begin(), dm.end(), gap) - dm.begin()) - 1;
if(p == -1){
// gap smaller than any dm[0]
if(aqua[0] > 0){
ans[i] += min(aqua[0], gap);
gap -= min(aqua[0], gap);
if(gap > 0) ans[i+1] += gap;
} else {
ans[i+1] += min(llabs(aqua[0]), gap);
gap -= min(llabs(aqua[0]), gap);
if(gap > 0) ans[i] += gap;
}
}
else if(p == (int)dm.size()-1){
// use the very last “flip”
if(aqua[p] > 0){
ans[i] += aqua[p];
ans[i+1] += llabs(aqua[p+1]);
} else {
ans[i] += aqua[p+1];
ans[i+1] += llabs(aqua[p]);
}
}
else {
// take dm[p] fully, then split the rest
if(aqua[p] > 0){
ans[i] += aqua[p] + (gap - dm[p]);
ans[i+1] += llabs(aqua[p+1]);
} else {
ans[i] += aqua[p+1];
ans[i+1] += llabs(aqua[p]) + (gap - dm[p]);
}
}
}
// 6) add the final dangling bits
ans[1] += bestNeg;
ans[n] += bestPos;
// 7) print
for(int i = 1; i <= n; i++){
cout << ans[i] << "\n";
}
return 0;
}
