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;

const int N = 2e5;
const int inf = 1 << 30;
int ans[N], ds[N];

int main(void){
  typedef tuple<int, int, int> t3;
  int n, Q;
  priority_queue<t3, vector<t3>, greater<t3>> q;
  cin >> n >> Q;
  multiset<int> ms; ms.insert(inf);
  
  for (int i = 0; i < n; i++){
    int s, t, x;
    scanf("%d %d %d", &s, &t, &x);
    q.push(make_tuple(s - x, x, 1));
    q.push(make_tuple(t - x, x, 0));
  }

  for (int i = 0; i < Q; i++) scanf("%d", &ds[i]), ans[i] = -1;

  int cur = 0;
  while (!q.empty()){
    t3 e = q.top(); q.pop();
    int t, x, act, v;
    tie(t, x, act) = e; v = *ms.begin();
    while (cur < Q && ds[cur] < t)
      ans[cur++] = v;
    
    if (act) ms.insert(x);
    else ms.erase(ms.find(x));
  }

  for (int i = 0; i < Q; i++) printf("%d\n", ans[i] == inf ? -1 : ans[i]);
  return 0;
}
