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++14 (GCC 5.4.1) (judge verdict: Accepted)

/***********Template Starts Here***********/
#include <bits/stdc++.h>

#define pb push_back
#define nl puts ("")
#define sp printf ( " " )
#define phl printf ( "hello\n" )
#define ff first
#define ss second
#define POPCOUNT __builtin_popcountll
#define RIGHTMOST __builtin_ctzll
#define LEFTMOST(x) (63-__builtin_clzll((x)))
#define MP make_pair
#define FOR(i,x,y) for(vlong i = (x) ; i <= (y) ; ++i)
#define ROF(i,x,y) for(vlong i = (y) ; i >= (x) ; --i)
#define CLR(x,y) memset(x,y,sizeof(x))
#define UNIQUE(V) (V).erase(unique((V).begin(),(V).end()),(V).end())
#define MIN(a,b) ((a)<(b)?(a):(b))
#define MAX(a,b) ((a)>(b)?(a):(b))
#define NUMDIGIT(x,y) (((vlong)(log10((x))/log10((y))))+1)
#define SQ(x) ((x)*(x))
#define ABS(x) ((x)<0?-(x):(x))
#define FABS(x) ((x)+eps<0?-(x):(x))
#define ALL(x) (x).begin(),(x).end()
#define LCM(x,y) (((x)/gcd((x),(y)))*(y))
#define SZ(x) ((vlong)(x).size())
#define NORM(x) if(x>=mod)x-=mod;
#define MOD(x,y) (((x)*(y))%mod)
#define ODD(x) (((x)&1)==0?(0):(1))

using namespace std;

typedef unsigned int uint;
typedef long long vlong;
typedef unsigned long long uvlong;
typedef pair<int, int> pii;
typedef pair < vlong, vlong > pll;
typedef vector<pii> vii;
typedef vector<pll> vll;
typedef vector<int> vi;
typedef vector<vlong> vl;
using ll = long long;

const vlong inf = 2147383647;
const double pi = 2 * acos ( 0.0 );
const double eps = 1e-9;
mt19937 rng(std::chrono::duration_cast<std::chrono::nanoseconds>(chrono::high_resolution_clock::now().time_since_epoch()).count());

template<typename S, typename T>
void xmin(S&a, T const&b){if(b<a) a=b;}
template<typename S, typename T>
void xmax(S&a, T const&b){if(b>a) a=b;}

#ifdef DEBUG
     clock_t tStart = clock();
     #define debug(args...) {dbg,args; cerr<<endl;}
     #define timeStamp debug ("Execution Time: ", (double)(clock() - tStart)/CLOCKS_PER_SEC)
     #define bug printf("%d\n",__LINE__);

#else
    #define debug(args...)  // Just strip off all debug tokens
    #define timeStamp
#endif

struct debugger{
    template<typename T> debugger& operator , (const T& v){
        cerr<<v<<" ";
        return *this;
    }
}dbg;

//int knightDir[8][2] = { {-2,1},{-1,2},{1,2},{2,1},{2,-1},{-1,-2},{1,-2},{-2,-1} };
//int dir4[4][2] = {{-1,0},{0,1},{1,0},{0,-1}};

inline vlong gcd ( vlong a, vlong b ) {
    a = ABS ( a ); b = ABS ( b );
    while ( b ) { a = a % b; swap ( a, b ); } return a;
}

vlong ext_gcd ( vlong A, vlong B, vlong *X, vlong *Y ){
    vlong x2, y2, x1, y1, x, y, r2, r1, q, r;
    x2 = 1; y2 = 0;
    x1 = 0; y1 = 1;
    for (r2 = A, r1 = B; r1 != 0; r2 = r1, r1 = r, x2 = x1, y2 = y1, x1 = x, y1 = y ) {
        q = r2 / r1;
        r = r2 % r1;
        x = x2 - (q * x1);
        y = y2 - (q * y1);
    }
    *X = x2; *Y = y2;
    return r2;
}

inline vlong modInv ( vlong a, vlong m ) {
    vlong x, y;
    ext_gcd( a, m, &x, &y );
    x %= m;
    if ( x < 0 ) x += m; //modInv is never negative
    return x;
}

inline vlong power ( vlong a, vlong p ) {
    vlong res = 1, x = a;
    while ( p ) {
        if ( p & 1 ) res = ( res * x );
        x = ( x * x ); p >>= 1;
    }
    return res;
}

inline vlong bigmod ( vlong a, vlong p, vlong m ) {
    vlong res = 1 % m, x = a % m;
    while ( p ) {
        if ( p & 1 ) res = ( res * x ) % m;
        x = ( x * x ) % m; p >>= 1;
    }
    return res;
}

/***********Template Ends Here***********/

signed gen(int T){
    // mt19937 rng(43151);
    //auto get_rand = [&](int64_t l, int64_t r){ return uniform_int_distribution<int64_t>(l, r)(rng); };
    //auto get_double = [&](double l, double r){ return uniform_real_distribution<double>(l, r)(rng); };
    ofstream o("gen.txt");
    o << T << "\n";
    for(int cas=0;cas<T;++cas){
        o << "\n";
        /// GEN HERE
        
    }
    o << endl;
    o.close();
    return 0;
}

signed main()
{
    #ifdef LOCAL_RUN
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    cin.tie(0); cout.tie(0); ios_base::sync_with_stdio(false);
    int TTT; cin >> TTT; 
	if(TTT < 0) return gen(-TTT);
	while(TTT--){
    #else
    cin.tie(0); cout.tie(0); ios_base::sync_with_stdio(false);
    #endif // LOCAL_RUN

    ///CODE
    int n, q;
    cin >> n >> q;
    vi S(n), T(n), X(n);
    FOR(i, 0, n - 1) {
        cin >> S[i] >> T[i] >> X[i];
        S[i] -= X[i];
        T[i] -= X[i];
    }
    vi id(n);
    iota(ALL(id), 0);
    sort(ALL(id), [&](int i, int j) {
        return X[i] < X[j];
    });
    map<int, int> times;
    FOR(i, 0, q - 1) {
        int t;
        cin >> t;
        times[t] = i;
    }
    vi res(q, -1);
    FOR(i, 0, n - 1) {
        int idx = id[i], s = S[idx], t = T[idx];
        while (not times.empty()) {
            auto it = times.lower_bound(s);
            if (it == times.end() or it->first >= t) {
                break;
            }
            int x = it->second;
            res[x] = X[idx];
            times.erase(it);
        }
    }
    for (int x: res) {
        cout << x << "\n";
    }

    #ifdef LOCAL_RUN
    }
    #endif // LOCAL_RUN
    return 0;
}