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;
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;
typedef pair<ll,ll> pll;
const ll mod=1e9+7;
//const ll mod=998244353;
const ll inf=1e18;

vector<pair<pll,ll>> E;

int main() {
  ll n,q;
  cin >> n >> q;
  for(ll i=0;i<n;i++) {
    ll s,t,x;
    cin >> s >> t >> x;
    s-=x;
    t-=x;
    E.push_back({{s,t},x});
  }
  ll id=0;
  sort(E.begin(),E.end());
  priority_queue<pll> p;
  for(ll i=0;i<q;i++) {
    ll d;cin >> d;
    while(id<n&&E[id].first.first<=d) {
      p.push({-E[id].second,E[id].first.second});
      id++;
    }
    while(p.size()&&p.top().second<=d) p.pop();
    if(p.size()) cout << -p.top().first << endl;
    else cout << -1 << endl;
  }
}