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<bits/stdc++.h>
using namespace std;
#define FOR(i,n) for(int i=0;i<n;i++)
#define FORX(i,n) for(int i=1;i<=n;i++) 
#define scan(x) scanf("%d",&x)
#define print(x) printf("%d\n",x)
const int N=2e5+200;
struct node{
	int s,t,x;
}a[N];
int ans[N],n,m,value;
set<pair<int,int> >vis;
int cmp(const node &a,const node &b){
	return a.x<b.x;
}
int main(){
	scan(n);scan(m);
	FOR(i,n){
		scan(a[i].s);scan(a[i].t);scan(a[i].x);
		a[i].s-=a[i].x;a[i].t-=a[i].x;
	}
	sort(a,a+n,cmp); 
	memset(ans,-1,sizeof(ans));
	FOR(i,m){
	scan(value);
	vis.insert({value,i});}
	FOR(i,n){
		set<pair<int,int> >::iterator it;
		it=vis.lower_bound({a[i].s,0});
		while(it!=vis.end()){
			if(it->first>=a[i].t)break;
			ans[it->second]=a[i].x;
			vis.erase(it++);
		}
	}
	FOR(i,m) print(ans[i]);
}