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;
    vector<tuple<int, int, int>> a(n);
    for (int i = 0; i < n; i++) {
        int s, t, x; cin >> s >> t >> x;
        a[i] = make_tuple(x, s - x, t - x);
    }
    sort(a.begin(), a.end());
    set<pair<int, int>> st;
    for (int i = 0; i < q; i++) {
        int d; cin >> d;
        st.insert({d,i});
    }
    vector<int> ans(q, -1);
    for (int i = 0; i < n; i++) {
        int s = get<1>(a[i]);
        auto t = st.lower_bound({s,-1});
        while (t != st.end()) {
            if (t->first < get<2>(a[i])) {
                ans[t->second] = get<0>(a[i]);
                st.erase(t++);
            }
            else {
                break;
            }
        }
    }
    for (int k : ans) {
        cout << k << endl;
    }
    return 0;
}