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;
typedef long long ll;

int main(){
	int N, Q; cin >> N >> Q;
	int D[Q];
	vector<pair<int, pair<int, int>>> ve;
	for(int i=0; i<N; i++){
		int s, t, x; cin >> s >> t >> x;
		ve.push_back({x, {s, t}});
	}
	sort(ve.begin(), ve.end());
	set<int> st;
	for(int i=0; i<Q; i++){
		cin >> D[i];
		st.insert(D[i]);
	}
	map<ll, ll> ans;
	for(int i=0; i<N; i++){
		int p = ve[i].second.first - ve[i].first;
		int q = ve[i].second.second - ve[i].first;
		auto j = st.lower_bound(p);
		while(*j < q && j != st.end()){
			ans[*j] = ve[i].first;
			j = st.erase(j);
		}
	}
	auto j = st.lower_bound(-1);
	while(*j <= 1e9 && j != st.end()){
		ans[*j] = -1;
		j = st.erase(j);
	}
	for(int i=0; i<Q; i++){
		cout << ans[D[i]] << endl;
	}
	return 0;
}
