Problem:
Score : 500 points

Problem Statement

There is an infinitely long street that runs west to east, which we consider as a number line.

There are N roadworks scheduled on this street.
The i-th roadwork blocks the point at coordinate X_i from time S_i - 0.5 to time T_i - 0.5.

Q people are standing at coordinate 0. The i-th person will start the coordinate 0 at time D_i, continue to walk with speed 1 in the positive direction and stop walking when reaching a blocked point.

Find the distance each of the Q people will walk.

Constraints

All values in input are integers.

1 \leq N, Q \leq 2 \times 10^5

0 \leq S_i < T_i \leq 10^9

1 \leq X_i \leq 10^9

0 \leq D_1 < D_2 < ... < D_Q \leq 10^9

If i \neq j and X_i = X_j, the intervals [S_i, T_i) and [S_j, T_j) do not overlap.

Input

Input is given from Standard Input in the following format:

N Q
S_1 T_1 X_1
:
S_N T_N X_N
D_1
:
D_Q

Output

Print Q lines. The i-th line should contain the distance the i-th person will walk or -1 if that person walks forever.

Sample Input 1

4 6
1 3 2
7 13 10
18 20 13
3 4 2
0
1
2
3
5
8

Sample Output 1

2
2
10
-1
13
-1

The first person starts coordinate 0 at time 0 and stops walking at coordinate 2 when reaching a point blocked by the first roadwork at time 2.

The second person starts coordinate 0 at time 1 and reaches coordinate 2 at time 3. The first roadwork has ended, but the fourth roadwork has begun, so this person also stops walking at coordinate 2.

The fourth and sixth persons encounter no roadworks while walking, so they walk forever. The output for these cases is -1.

Solution in C++ (GCC 9.2.1) (judge verdict: Runtime Error)

//インクルード(アルファベット順)
#include<algorithm>//sort,二分探索,など
#include<bitset>//固定長bit集合
#include<cmath>//pow,logなど
#include<complex>//複素数
#include<deque>//両端アクセスのキュー
#include<functional>//sortのgreater
#include<iomanip>//setprecision(浮動小数点の出力の誤差)
#include<iostream>//入出力
#include<iterator>//集合演算(積集合,和集合,差集合など)
#include<map>//map(辞書)
#include<numeric>//iota(整数列の生成),gcdとlcm(c++17)
#include<queue>//キュー
#include<set>//集合
#include<stack>//スタック
#include<string>//文字列
#include<unordered_map>//イテレータあるけど順序保持しないmap
#include<unordered_set>//イテレータあるけど順序保持しないset
#include<utility>//pair
#include<vector>//可変長配列

using namespace std;
typedef long long ll;

//マクロ
//forループ関係
//引数は、(ループ内変数,動く範囲)か(ループ内変数,始めの数,終わりの数)、のどちらか
//Dがついてないものはループ変数は1ずつインクリメントされ、Dがついてるものはループ変数は1ずつデクリメントされる
#define REP(i,n) for(ll i=0;i<(ll)(n);i++)
#define REPD(i,n) for(ll i=n-1;i>=0;i--)
#define FOR(i,a,b) for(ll i=a;i<=(ll)(b);i++)
#define FORD(i,a,b) for(ll i=a;i>=(ll)(b);i--)
//xにはvectorなどのコンテナ
#define ALL(x) x.begin(),x.end() //sortなどの引数を省略したい
#define SIZE(x) ll(x.size()) //sizeをsize_tからllに直しておく
//定数
#define INF 1000000000000 //10^12:極めて大きい値,∞
#define MOD 1000000007 //10^9+7:合同式の法
#define MAXR 100000 //10^5:配列の最大のrange(素数列挙などで使用)
//略記
#define PB push_back //vectorヘの挿入
#define MP make_pair //pairのコンストラクタ
#define F first //pairの一つ目の要素
#define S second //pairの二つ目の要素
#define Umap unordered_map
#define Uset unordered_set


signed main(){
    //入力の高速化用のコード
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    ll n,q;cin>>n>>q;

    vector<vector<ll>> xst(n);
    REP(i,n){
        ll s,t,x;cin>>s>>t>>x;
        xst[i]={x,s,t};
    }
    sort(ALL(xst));

    vector<ll> d(q);
    REP(i,q)cin>>d[i];

    vector<deque<ll>> ans_sectl(q);
    vector<deque<ll>> ans_sectr(q);
    REP(i,n){
        ll s,t,x;x=xst[i][0];s=xst[i][1];t=xst[i][2];
        ll L=(lower_bound(ALL(d),s-x)-d.begin());
        ll R=(upper_bound(ALL(d),t-x-1)-d.begin())-1;
        ans_sectl[L].PB(i);ans_sectr[R].PB(i);
        //cout << L << " " << R << endl;
    }

    set<ll> ans;
    REP(i,q){
        ll sl=SIZE(ans_sectl[i]);
        ll sr=SIZE(ans_sectr[i]);
        REP(_,sl){
            ans.insert(*(ans_sectl[i].begin()));
            ans_sectl[i].pop_front();
        }
        if(ans.empty())cout<<-1<<"\n";
        else cout<<xst[*(ans.begin())][0]<<"\n";
        REP(_,sr){
            ans.erase(*(ans_sectr[i].begin()));
            ans_sectr[i].pop_front();
        }
    }
}