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 ll long long int

using namespace std;

const ll mod=1000000007;
const ll inf=1e18;

int main(){

  ll n,q;
  cin >> n >> q;

  ll s[n],t[n],x[n];
  vector<tuple<ll,ll,ll>> eve;//イベントを入れる配列
  for(int i=0;i<n;i++){
    cin >> s[i] >> t[i] >> x[i];
    eve.push_back(make_tuple(s[i]-x[i],1,x[i]));
    eve.push_back(make_tuple(t[i]-x[i],-1,x[i]));
  }

  for(int i=0;i<q;i++){
    ll d;
    cin >> d;
    eve.push_back(make_tuple(d,2,i));
  }

  sort(eve.begin(),eve.end());

  multiset<ll> mc;//現時点で通行止めにされている座標を持つセット

  for(int p=0;p<eve.size();p++){
    ll x=get<0>(eve[p]);
    ll y=get<1>(eve[p]);
    ll z=get<2>(eve[p]);
    if(y==1) mc.insert(z);
    else if(y==-1) mc.erase(mc.lower_bound(z));
    else{
      cout << (mc.empty() ? -1 : *(mc.begin())) << endl;
    }
  }



}