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>
//g++ -std=c++11 

using namespace std;
typedef long long ll;

#define rep(i,n) for (long long i = 0; i < (n); ++i)
#define DIV 1000000007 //10^9+7

ll ans[200005];

int main(){
	ll N, Q;
	cin >> N >> Q;

	vector<tuple<ll, ll, ll> > XST;

	rep(i, N) {
		ll s, t, x;
		cin >> s >> t >> x;
		s-=x;
		t-=x;
		XST.emplace_back(x, s, t);
	}

	sort(XST.begin(), XST.end());
	
	map<ll, ll> memo;
	set<ll> D;
	rep(i, Q) {
		ll tmp;
		cin >> tmp;
		D.insert(tmp);
		memo[tmp] = i;
		ans[i] = -1;
	}

	rep(i, N) {
		ll x = get<0>(XST[i]);
		ll s = get<1>(XST[i]);
		ll t = get<2>(XST[i]);

		set<ll>::iterator it = D.lower_bound(s);
		while(it != D.end() && *it < t) {
			ans[memo[*it]] = x;
			D.erase(it++);
		}
	}

	rep(i, Q) {
		cout << ans[i] << endl;
	}
}
