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,a) for(int i = 0; i < (a); i++)
#define ALL(a) (a).begin(),(a).end()
typedef long long ll;
typedef pair<ll, ll> P;
const int INF = 1e9;
const long long LINF = 1e18;
const long long MOD = 1e9 + 7;

signed main(){
    int n,q;
    cin >> n >> q;
    set<P> st;
    ll s,t,x;
    REP(i,n){
        cin >> s >> t >> x;
        s -= x;
        t -= x;
        st.insert(P(s, x));
        st.insert(P(t, -x));
    }
    ll ans[q];
    ll d;
    set<ll> now;
    auto itr = st.begin();
    REP(i,q){
        cin >> d;
        while(itr->first <= d && itr != st.end()){
            if(itr->second > 0) now.insert(itr->second);
            else now.erase(-(itr->second));
            itr++;
        }
        if(!now.empty()) ans[i] = *now.begin();
        else ans[i] = -1;
    }
    REP(i,q){
        cout << ans[i] << endl;
    }
    return 0;
}