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;
#define rep(i,n) for(int i=0;i<(n);i++)

int main(int argc, char const *argv[])
{
	int n, q; cin >> n >> q;
	vector<int> s(n),t(n),x(n); rep(i, n) cin >> s[i] >> t[i] >> x[i];
	vector<int> d(q); rep(i, q) cin >> d[i];

	set<pair<int, int>> st; rep(i, q) st.insert(make_pair(d[i], i));

	vector<tuple<int, int, int>> v;
	rep(i, n) v.push_back(make_tuple(x[i], s[i] - x[i],t[i] - x[i]));
	sort(v.begin(), v.end());

	vector<int> ans(q, -1);
	rep(i, n) {
		int x, l, r; tie(x, l, r) = v[i];
		auto it = st.lower_bound(make_pair(l, -1)); // O(log q)
		while (it != st.end()) {
			if (it->first < r) {
				ans[it->second] = x;
				st.erase(it++);
			} else {
				break;
			}
		}
	}

	for (auto u : ans) {
		printf("%d\n", u);
	}

	return 0;
}
