#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<(b);++i)
#define all(n) (n).begin(),(n).end()
using lint=long long;

/**
 * Author: Simon Lindholm
 * Date: 2017-04-20
 * License: CC0
 * Source: own work
 * Description: Container where you can add lines of the form kx+m, and query maximum values at points x.
 *  Useful for dynamic programming (``convex hull trick'').
 * Time: O(\log N)
 * Status: stress-tested
 */

using ll=long long;
struct Line {
	mutable ll k, m, p;
	bool operator<(const Line& o) const { return k < o.k; }
	bool operator<(ll x) const { return p < x; }
};

struct LineContainer : multiset<Line, less<>> {
	// (for doubles, use inf = 1/.0, div(a,b) = a/b)
	static const ll inf = LLONG_MAX;
	ll div(ll a, ll b) { // floored division
		return a / b - ((a ^ b) < 0 && a % b); }
	bool isect(iterator x, iterator y) {
		if (y == end()) return x->p = inf, 0;
		if (x->k == y->k) x->p = x->m > y->m ? inf : -inf;
		else x->p = div(y->m - x->m, x->k - y->k);
		return x->p >= y->p;
	}
	void add(ll k, ll m) {
		auto z = insert({k, m, 0}), y = z++, x = y;
		while (isect(y, z)) z = erase(z);
		if (x != begin() && isect(--x, y)) isect(x, y = erase(y));
		while ((y = x) != begin() && (--x)->p >= y->p)
			isect(x, erase(y));
	}
	ll query(ll x) {
		assert(!empty());
		auto l = *lower_bound(x);
		return l.k * x + l.m;
	}
};

int main(){
    cin.tie(0)->sync_with_stdio(0);
    lint n,q;
    cin>>n>>q;
    vector<lint>a(n);
    rep(i,0,n)cin>>a[i];
    map<lint,lint>memo;
    LineContainer lc;
    rep(i,0,n)lc.add(-i,a[i]);
    lint sum=0;
    rep(i,0,n)sum+=a[i];
    vector<lint>dd(n-1);
    rep(i,1,n)dd[i-1]+=a[i]-a[i-1];
    sort(all(dd));
    vector<lint>dsum(n);
    rep(i,0,n-1)dsum[i+1]=dd[i]+dsum[i];
    while(q--){
        lint d;
        cin>>d;
        if(memo.count(d)){
            cout<<memo[d]<<endl;
            continue;
        }
        lint mn=1LL<<60;
        lint x=lc.query(d);
        auto e=lower_bound(all(dd),d)-dd.begin();
        lint ans=(d*e-dsum[e])+((dsum.back()-dsum[e])-d*(n-1-e));
        cout<<(ans+abs(x+(n-1)*d-a.back())+abs(a[0]-x))/2<<endl;
    }
}