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;
int N, Q;
int S[202020], T[202020], X[202020];
int main()
{
  const int INF = 2e9+10;
  scanf("%d%d", &N, &Q);
  vector<pair<int, int> > V;
  for(int i=0; i<N; ++i)
  {
    scanf("%d%d%d", S+i, T+i, X+i);
    V.emplace_back(S[i]-X[i], X[i]);
    V.emplace_back(T[i]-X[i], -X[i]);
  }
  for(int i=0; i<Q; ++i)
  {
    int t; scanf("%d", &t);
    V.emplace_back(t, INF);
  }
  multiset<int> M;
  sort(V.begin(), V.end());
  for(auto x: V)
  {
    int a = x.first, b = x.second;
    if(b<0) M.erase(M.find(-b));
    else if(b==INF)
    {
      if(M.size() == 0) printf("%d\n", -1);
      else printf("%d\n", *M.begin());
    }
    else M.insert(b);
  }
  
}