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 + 10;
int n, q, S[N], T[N], X[N], R[N], P[N], D[N];
bool CMP(int i, int j)
{
    return (X[i] < X[j]);
}
int main()
{
    scanf("%d%d", &n, &q);
    for (int i = 0; i < n; i++)
        scanf("%d%d%d", &S[i], &T[i], &X[i]), P[i] = i;
    set < pair < int , int > > ST;
    for (int i = 0; i < q; i++)
    {
        scanf("%d", &D[i]);
        ST.insert({D[i], i});
        R[i] = -1;
    }
    sort(P, P + n, CMP);
    for (int j = 0; j < n; j++)
    {
        int i = P[j];
        auto it = ST.lower_bound(pair < int , int > {S[i] - X[i], -1});
        while (it != ST.end() && it->first + X[i] < T[i])
            R[it->second] = X[i], it = ST.erase(it);
    }
    for (int i = 0; i < q; i++)
        printf("%d\n", R[i]);
    return 0;
}
