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 Rust (1.15.1) (judge verdict: Wrong Answer)

macro_rules! get {
    (Vec<$t:ty>) => {
        {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            line.split_whitespace()
                .map(|t| t.parse::<$t>().unwrap())
                .collect::<Vec<_>>()
        }
    };
    ($t:ty) => {
        {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            line.trim().parse::<$t>().unwrap()
        }
    };
    ($($t:ty),*) => {
        {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            (
                $(iter.next().unwrap().parse::<$t>().unwrap(),)*
            )
        }
    };
    ($t:ty; $n:expr) => {
        (0..$n).map(|_|
            get!($t)
        ).collect::<Vec<_>>()
    };
    ($($t:ty),*; $n:expr) => {
        (0..$n).map(|_|
            get!($($t),*)
        ).collect::<Vec<_>>()
    };
}

use std::collections::BinaryHeap;
use std::collections::HashSet;

fn main() {
    let (N, Q) = get!(usize, usize);
    let STX = get!(i64, i64, i64; N);
    let D = get!(i64; Q);

    let INF: i64 = 1000000000000000;

    // 各時間帯について、その時間帯に出発すると影響を受ける工事を前もって求める
    // 前もって求めた時間帯→工事を用いて、各クエリに回答する
    let mut C = vec!();
    for (s, t, x) in STX {
        C.push((s - x, x, 1));
        C.push((t - x, x, 0));
    }
    C.sort();
    let mut q = BinaryHeap::new();
    let mut R = HashSet::new();
    let mut u = Vec::new();
    let mut v = Vec::new();
    q.push(-INF);
    for (t, x, k) in C {
        let p = -q.peek().unwrap();
        if k == 1 {
            if x < p {
                u.push(t);
                v.push(p);
            }
            q.push(-x);
        } else {
            R.insert(x);
            if x == p {
                u.push(t);
                v.push(p);
            }
            while R.contains(&-q.peek().unwrap()) {
                R.remove(&-q.pop().unwrap());
            }
        }
    }
    u.push(INF);
    v.push(INF);
    let mut Y = Vec::new();
    let mut i = 0;
    for d in D {
        while u[i] <= d {
            i += 1;
        }
        Y.push(v[i]);
    }
    for y in Y {
        if y == INF {
            println!("-1");
        } else {
            println!("{}", y)
        }
    }
}