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;
const int N = 2e5+87, M=1e9+7;
int s[N],t[N],x[N],d[N],ans[N];
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n,q;
    cin >> n >> q;
    vector<pair<int,int>> e;
    for (int i = 0; i < n; ++i) {
        cin >> s[i] >> t[i] >> x[i];
        s[i] -= x[i];
        t[i] -= x[i];
        e.emplace_back(s[i],x[i]);
        e.emplace_back(t[i],-x[i]);
    }
    for (int i = 0; i < q; ++i) {
        cin >> d[i];
        e.emplace_back(d[i], M+i);
    }
    sort(begin(e),end(e));
    multiset<int> st{M};
    for (auto p : e) {
        int x = p.second;
        if (x >= M) {
            ans[x-M] = *begin(st);
        } else if (x > 0) {
            st.insert(x);
        } else {
            st.erase(st.lower_bound(-x));
        }
    }
    for (int i = 0; i < q; ++i)
        cout << (ans[i] == M ? -1 : ans[i]) << '\n';
}
