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;
#define REP(i, n) for(int i=0; i<(int)(n); i++)
int main(){
int N, Q, S, T, X, D; cin >> N >> Q; vector<int> ans(Q, -1);
vector< vector<int> > A(N);
REP(i, N){ cin >> S >> T >> X; A[i]={X, S, T}; }
sort(A.begin(), A.end());
set< pair<int, int> > P;
REP(i, Q){ cin >> D; P.insert(make_pair(D, i)); }
REP(i, N){
  int x=A[i][0], s=A[i][1], t=A[i][2], j=0;
  pair<int, int> l=make_pair(s-x, -1), r=make_pair(t-x, -1);
  auto L=P.lower_bound(l), R=P.lower_bound(r);
  vector< pair<int, int> > K;
  while(L!=R){ j=(*L).second; ans[j]=x; K.push_back(*L); L++; }
  REP(k, K.size()){ P.erase(K[k]); }
}
REP(i, Q){ cout << ans[i] << "\n"; }
}