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>
#define rep(i,n) for(int i=0; i<(n); i++)
using namespace std;
using ll = long long;

int main() 
{
    int n, q;
    cin >> n >> q;

    vector<tuple<int,int,int>> V;
    rep(i,n) {
        int l, r, x;
        cin >> l >> r >> x;
        V.push_back(forward_as_tuple(x, l-x, r-x));
    }
    sort(V.begin(), V.end());

    set<pair<int,int>> S;
    rep(i,q) {
        int d;
        cin >> d;
        S.insert(make_pair(d,i));
    }

    int ans[q];
    rep(i,q) ans[i] = -1;

    for (auto v : V) {
        int x, l, r;
        tie(x, l, r) = v;
        auto it = S.lower_bound(make_pair(l,0));
        while ( it != S.end()) {
            int d = it->first;
            int idx = it->second;
            if (d >= r) break;
            ans[idx] = x;
            S.erase(it++);
        }
    }

    rep(i,q) cout << ans[i] << endl;

    return 0;
}
