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
import sys
input = sys.stdin.readline
import bisect
n, q = map(int, input().split())

ran = []
for i in range(n):
    s, t, x = map(int, input().split())
    ran.append((s-x, t-x-1, x))
ran.sort(key=lambda a: a[2])

qu = []
convert = {}
for i in range(q):
    que = int(input())
    convert[que] = i
    qu.append(que)
qu.sort()

memo = [-1] * q

ans = [-1] * q

for i in range(n):
    x = ran[i][2]
    l = ran[i][0]
    r = ran[i][1]
    ql = bisect.bisect_left(qu, l)
    qr = bisect.bisect_right(qu, r)
    if ql >= qr:
        continue
    while ql < qr:
        if memo[ql] >= 0:
            ql = memo[ql]
        else:
            memo[ql] = qr
            ans[convert[qu[ql]]] = x
            ql += 1

for i in ans:
    print(i)