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 int ll;
#define rep(i, srt, end) for (long long i = (srt); i < (long long)(end); i++)

typedef struct shirakami {
    ll pos, c, x;
} fbk;

int main(){
    ll n, q;
    cin >> n >> q;
    vector<fbk> a(2*n);
    rep(i, 0, n){
        ll s, t, x;
        cin >> s >> t >> x;
        a[2*i] = {s-x, 1, x};
        a[2*i+1] = {t-x, 0, x};
    }
    vector<ll> d(q);
    rep(i, 0, q) cin >> d[i];
    sort(a.begin(), a.end(), [](fbk lhs, fbk rhs) {
        return (lhs.pos < rhs.pos);
    });
    ll k = 0;
    multiset<ll> st;
    rep(i, 0, q) {
        for(; k < 2*n && a[k].pos <= d[i]; k++) {
            if(a[k].c) st.insert(a[k].x);
            else st.erase(st.find(a[k].x)); 
        }
        if(st.empty()) cout << -1 << endl;
        else cout << *st.begin() << endl;
    }
    return 0;
}