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.

Example

Input

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


Output

2
2
10
-1
13
-1
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#define N 200005
using namespace std;

struct st {int s, t, x;} a[N];
int n, q, d;
priority_queue<st> pq;

bool operator < (st p, st q) {return p.x > q.x;}
bool cmp(st p, st q) {
	if(p.s != q.s) return p.s < q.s;
	else return p.t < q.t;
}

int main() {
	int i;
	cin >> n >> q;
	for(i=0; i<n; i++) {
		scanf("%d%d%d", &a[i].s, &a[i].t, &a[i].x);
		a[i].s -= a[i].x;
		a[i].t -= a[i].x;
	}
	sort(a, a+n, cmp);
	i = 0;
	while(q--) {
		scanf("%d", &d);
		for(; i<n && a[i].s<=d; i++) pq.push(a[i]);
		while(!pq.empty()) {
			st z = pq.top();
			if(z.t > d) {printf("%d\n", z.x); break;}
			pq.pop();
		}
		if(pq.empty()) puts("-1");
	}
	return 0;
}