//yukicoder@cpp17

#include <iostream>
#include <iomanip>

#include <algorithm>

#include <cmath>
#include <cctype>
#include <climits>
#include <cassert>

#include <string>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#include <map>

#include <random>

#include <bitset>

#include <complex>

#include <utility>

#include <numeric>

#include <functional>


using namespace std;
using ll = long long;
using P = pair<ll,ll>;

const ll MOD = 998244353;
const ll MODx = 1000000007;
const int INF = (1<<30)-1;
const ll LINF = (1LL<<62LL)-1;
const double EPS = (1e-10);

P ar4[4] = {{0,1},{0,-1},{1,0},{-1,0}};
P ar8[8] = {{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};




template <typename T> vector<T> make_vector(size_t a, T b) { return vector<T>(a, b); }
template <typename... Ts> auto make_vector(size_t a, Ts... ts) { return vector<decltype(make_vector(ts...))>(a, make_vector(ts...)); }
 
/*
確認ポイント
cout << fixed << setprecision(n) << 小数計算//n桁の小数表記になる

計算量は変わらないが楽できるシリーズ
min(max)_element(iter,iter)で一番小さい(大きい)値のポインタが帰ってくる
count(iter,iter,int)でintがiterからiterの間にいくつあったかを取得できる
*/

/* comment outed because can cause bugs
__attribute__((constructor))
void initial() {
  cin.tie(0);
  ios::sync_with_stdio(false);
}
*/

int main(){
  int n,q;cin>>n>>q;
  vector<pair<int,long long>> A(n);
  vector<int> B(n);
  for(int i = 0; n > i; i++){
    cin>>A[i].first>>A[i].second;
    B[i] = A[i].first;
  }
  B.push_back(-1);
  A.push_back({-1, 0});
  B.push_back(1000000001);
  A.push_back({1000000001, 0});
  sort(A.begin(), A.end());
  sort(B.begin(), B.end());

  vector<long long> rR(A.size(), 0);
  long long cur = A[0].second;
  for(int i = 1; A.size() > i; i++){
    rR[i] = rR[i-1] + abs(A[i].first-A[i-1].first)*cur;
    cur += A[i].second;
  }

  reverse(A.begin(), A.end());
  vector<long long> rL(A.size(), 0);
  cur = A[0].second;
  for(int i = 1; A.size() > i; i++){
    rL[i] = rL[i-1] + abs(A[i].first-A[i-1].first)*cur;
    cur += A[i].second;
  }
  
  reverse(rL.begin(), rL.end());
  reverse(A.begin(), A.end());
  //for(int i = 0; A.size() > i; i++){
  //  cout << rR[i] << " ";
  //}
  //cout << endl;
  //for(int i = 0; A.size() > i; i++){
  //  cout << rL[i] << " ";
  //}
  //cout << endl;
  for(int i = 0; q > i; i++){
    long long x;cin>>x;
    auto z = lower_bound(B.begin(), B.end(), x) - B.begin();

    long long ans = (rR[z]-rR[z-1])/(B[z]-B[z-1])*(x-B[z-1]) + rR[z-1] + (rL[z-1]-rL[z])/(B[z]-B[z-1])*(B[z]-x) + rL[z];
    cout << ans << endl;


  }
  return 0;
}