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++)
const int INF = 2e9;
using P = pair<int,int>;

int main(){
    int n, q;
    cin >> n >> q;
    int s[n], t[n], x[n];
    vector<P> eve;
    set<int> se;
    rep(i,n) {
      cin >> s[i] >> t[i] >> x[i];
      eve.push_back(P(s[i]-x[i],x[i]));
      eve.push_back(P(t[i]-x[i],-x[i]));
    }
    int d[q];
    rep(i,q) cin >> d[i];
    sort(eve.begin(),eve.end());
    int p = 0;
    int ans[q];
    rep(i,q){
      while(p < eve.size() && eve[p].first <= d[i]) {
        if(eve[p].second > 0) {
          se.insert(eve[p].second);
        } else {
          se.erase(-eve[p].second);
        }
        p++;
      }
      if(se.empty()) {
        ans[i] = -1;
      } else {
        ans[i] = *se.begin();
      }
    }
    for(auto k : ans) {
      cout << k << endl;
    }
    return 0;
}