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
def main():
    from bisect import bisect_left
    def update(l,r,v):
        L,R=l+N0,r+N0
        while L<R:
            if R&1:
                R-=1
                data[R-1]=v
            if L&1:
                data[L-1]=v
                L+=1
            L>>=1
            R>>=1
    def query(k):
        k+=N0-1
        s=10**18
        while k+1:
            if data[k]<s:s=data[k]
            k=~-k//2
        return s
    n,*t=open(0)
    n,q=map(int,n.split())
    d=[-10**18]+list(map(int,t[n:]))+[10**18]
    N0=2**(q+1).bit_length()
    data=[10**18]*2*N0
    for x,t,s in sorted(list(map(int,t.split()))[::-1]for t in t[:n])[::-1]:
        l=bisect_left(d,s-x)
        r=bisect_left(d,t-x)
        update(l,r,x)
    for i in range(q):
        a=query(i+1)
        print(-(a==10**18)or a)
if __name__=='__main__':
    main()