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>
typedef long long LL;
#define FOR(i,a,b) for(LL i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)

using namespace std;
LL p = 1000000007ll;

int main(void)
{
  LL n,q;
  cin >> n >> q;
  map<LL,LL> ans;
  ans[-p] = -1;
  vector<tuple<LL,LL,LL>> xst;
  REP(i,n) {
    LL s,t,x;
    cin >> s >> t >> x;
    xst.emplace_back(x,s,t);
  }
  sort(xst.rbegin(),xst.rend());
  for(auto tu:xst) {
    LL s,t,x;
    tie(x,s,t) = tu;
    auto ite = ans.upper_bound(t-x);
    --ite;
    LL prev = ite->second;
    ans[s-x] = x;
    ans.erase(ans.upper_bound(s-x), ans.lower_bound(t-x));
    if(!ans.count(t-x)) ans[t-x] = prev;
  }
  REP(i,q) {
    LL d;
    cin >> d;
    auto ite = ans.upper_bound(d);
    --ite;
    cout << ite->second << endl;
  }
  return 0;
}
