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;
int n, q, s, t, x, d, u, j;
typedef pair<int,int> P;
P pr[400005];
set<int> st;
int main(){
    scanf("%d%d", &n, &q);
    for(int i = 0; i < n; i++){
        scanf("%d%d%d", &s, &t, &x);
        pr[u++] = P{s-x, x};
        pr[u++] = P{t-x, x};
    }
    sort(pr, pr+u);
    for(int i = 0; i < q; i++){
        scanf("%d", &d);
        while(j < u && pr[j].first <= d){
            auto it = st.find(pr[j].second);
            if(it == st.end()) st.insert(pr[j].second);
            else st.erase(it);
            j++;
        }
        printf("%d\n", st.empty() ? -1 : *st.begin());
    }
}