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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

#define rep(i,n) for(ll (i)=0; (i)<(ll)(n); (i)++)
#define frep(i,m,n) for(ll (i)=(m); (i)<=(ll)(n); (i)++)
#define rrep(i,n) for(ll (i)=(n)-1; (i)>-1; (i)--)
#define frrep(i,m,n) for(ll (i)=(n); (i)>(ll)(m); (i)--)
#define ALL(x) (x).begin(), (x).end()
#define SQR(x) (x)*(x)
#define CEIL(a,b) ((a)+(b)-1) /(b)

const int INF = 1001001001;
const int MOD = 1000000007;
const int MAX_N = 101010;

// A:array of pair, N:size of array
void sort_second(vector<pair<ll, ll>> &A, int N);
// a^n
ll a_n(ll a, ll n);
// make vector of (n!)%MOD, (n!)^(-1)%MOD to calculate nCk
void factorial_map(vector<ll> &n, vector<ll> &ninv);
// a^(-1) % MOD 
ll inv(ll a);
// map of prime factorization
void prime_map(map<ll, ll> &a, ll n);
//list of prime (is_prime.size() = n + 1, primes.size() = 0)
void prime_list(ll n, vector<ll> &primes, vector<bool> &is_prime);
// get gcd from prime map
ll gcd_map(map<ll, ll> a, map<ll, ll> b);
// get lcm from prime map
ll lcm_map(map<ll, ll> a, map<ll, ll> b);
// get gcd
ll my_gcd(ll a, ll b);
// calculate nCk
ll nCk(ll n, ll k);
// get abs
ll my_abs(ll a);
// z-algorithm (s.size() == a.size())
void z_algorithm(string s, vector<int> &a);

class Unionfind {
    public:
        vector<ll> par, siz, rank;
        Unionfind();
        ll root(ll x);
        void merge(ll x, ll y);
        bool issame(ll x, ll y);
        ll size(ll x);
};

int main() {
    ll n, q;
    cin >> n >> q;
    vector<pair<ll, pair<ll, ll>>> work(n);
    set<ll> d;
    vector<ll> index(q); // 取り出した値とindexを合わせる
    vector<ll> ans(q, -1); // 回答用

    rep(i, n) {
        ll s, t, x;
        cin >> s >> t >> x;
        work[i].first = x;
        work[i].second.first = s;
        work[i].second.second = t;
    }
    sort(ALL(work));
    rep(i, q) {
        ll x;
        cin >> x;
        d.insert(x);
        index[i] = x;
    }

    rep(i, n) {
        auto start = lower_bound(ALL(d), work[i].second.first-work[i].first); // 含む
        auto end = lower_bound(ALL(d), work[i].second.second-work[i].first); // 含んではいけない

        for(auto j = start; j != end; j++) {
            ans[lower_bound(ALL(index), *j)-index.begin()] = work[i].first;
        }
        d.erase(start, end);
    }

    rep(i, q) cout << ans[i] << endl;

    return 0;
}

void sort_second(vector<pair<ll, ll>> &A, int N) {
    rep(i, N) swap(A.at(i).first, A.at(i).second);
    sort(ALL(A));
    rep(i, N) swap(A.at(i).first, A.at(i).second);
}
ll a_n(ll a, ll n) { 
    if(n == 0) return 1;
    
    ll ret = a, count = 1;
    while(count * 2 < n) {
        ret *= ret;
        ret %= MOD;
        count *= 2;
    }
    if(count == n) return ret;
    else return (ret * a_n(a, n-count))%MOD ;
}

/* calculate a^n with no recursive process (%MOD)
ll a_n(ll a, ll n) { 
    ll ret = a, count = 1;
    int num = 1;
    vector<ll> memory(100);
    memory.at(0) = a;

    while(count * 2 < n) {
        ret *= ret;
        ret %= MOD;
        count *= 2;
        memory.at(num) = ret;
        num++;
    }
    if(count == n) return ret;
    else {
        bitset<100> bit = bitset<100>(n-count);
        rep(i, 100) {
            if(bit.test(i)) ret = (ret * memory.at(i)) % MOD;
        }
        return ret;
    }    
}
*/

void factorial_map(vector<ll> &n, vector<ll> &ninv) {
    n[0] = 1;
    for(ll i = 1; i < n.size(); i++) n[i] = (n[i-1] * i) % MOD;
    ninv[n.size()-1] = inv(n[n.size()-1]);
    for(ll i = n.size() - 2; i > -1; i--) ninv[i] = (ninv[i+1] * (i+1)) % MOD;
}

ll inv(ll a) {
    return a_n(a, MOD-2);
}

void prime_map(map<ll, ll> &a, ll n) {
    while((n%2) == 0) {
        n /= 2;
        if(a.count(2)) a.at(2)++;
        else a[2] = 1;
    }
    for(ll i = 3; i*i <= n; i+=2) {
        while((n%i) == 0) {
            n /= i;
            if(a.count(i)) a.at(i)++;
            else a[i] = 1;
        }
    }
    if(n != 1) a[n] = 1;
}

ll gcd_map(map<ll, ll> a, map<ll, ll> b) {
    ll ret = 1;
    for(auto p : a) {
        auto key = p.first;
        auto value = p.second;
        if(b.count(key)) {
            if(value > b.at(key)) ret *= a_n(key, b.at(key));
            else ret *= a_n(key, value);
        }
    }
    return ret;
}

ll lcm_map(map<ll, ll> a, map<ll, ll> b) {
    ll ret = 1;
    for(auto p: a) {
        auto key = p.first;
        auto value = p.second;
        if(b.count(key)) {
            if(value > b.at(key)) ret *= a_n(key, value);
            else ret *= a_n(key, b.at(key));
        }
        else {
            ret *= a_n(key, value);
        }
    }
    for(auto p: b) {
        auto key = p.first;
        auto value = p.second;
        if(a.count(key)) {
            continue;
        }
        else {
            ret *= a_n(key, value);
        }
    }
    return ret;
}

ll my_gcd(ll a, ll b) {
    if(b == 0) return a;
    return my_gcd(b, a%b);
}

ll nCk(ll n, ll k) {
    ll kfac = 1, ret = 1;
    for(ll i = 1; i <= k; i++) {
        kfac = (kfac * i) % MOD;
    }
    for(ll i = n; i >= n-k+1; i--) {
        ret = (ret * i) % MOD;
    }
    ret = (ret * inv(kfac)) % MOD;
    return ret;
}

void prime_list(ll n, vector<ll> &primes, vector<bool> &is_prime) {
    ll p = 0;
    rep(i, n + 1) is_prime[i] = true;
    is_prime[0] = false;
    is_prime[1] = false;
    for(ll i = 2; i <= n; i++) {
        if(is_prime[i]) {
            primes.push_back(i);
            for(ll j = 2*i; j <= n; j += i) is_prime[j] = false;
        }
    }
}

ll my_abs(ll a) {
    if(a >= 0) return a;
    else return -1 *a;
}

Unionfind::Unionfind() {
    par.resize(MAX_N);
    siz.resize(MAX_N);
    rank.resize(MAX_N);
    rep(i, MAX_N) {
        par[i] = i;
        siz[i] = 1;
        rank[i] = 0;
    }
}
ll Unionfind::root(ll x) {
    if(par[x] == x) return x;
    else {
        return par[x] = root(par[x]);
    }
}
void Unionfind::merge(ll x, ll y) {
    x = root(x);
    y = root(y);
    if(x == y) return;
    if(rank[x] < rank[y]) {
        par[x] = y;
        siz[y] += siz[x];
    }
    else {
        par[y] = x;
        siz[x] += siz[y];
        if(rank[x] == rank[y]) rank[x]++;
    }
}
bool Unionfind::issame(ll x, ll y) {
    return root(x) == root(y);
}
ll Unionfind::size(ll x) {
    return siz[root(x)];
}

void z_algorithm(string s, vector<int> &a) {
    a.at(0) = s.size();
    int i = 1, j = 0;
    while(i < s.size()) {
        while((i+j < s.size()) && (s.at(j) == s.at(i+j))) j++;
        a.at(i) = j;
        if(j == 0) {
            i++;
            continue;
        }
        int k = 1;
        while ((i+k < s.size()) && (k+a.at(k) < j)) {
            a.at(i+k) = a.at(k);
            k++;
        }
        i += k;
        j -= k;
    }
}