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;
#define N 200010
#define ll long long
const ll mod=1e9+7;

struct node
{
    int s,t,x;
    bool operator < (const node& b)const
    {
        return x>b.x;
    }
}a[N];
bool cmp(node a,node b)
{
    return a.s<b.s;
}
priority_queue<node>q;
int d[N];
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d%d",&a[i].s,&a[i].t,&a[i].x);
        a[i].s-=a[i].x,a[i].t-=a[i].x;
    }
    for(int i=1;i<=m;i++)scanf("%d",&d[i]);
    sort(a+1,a+1+n,cmp);
    int t=1,flag;
    for(int i=1;i<=m;i++)
    {
        flag=0;
        while(d[i]>=a[t].s && t<=n)
            q.push(a[t++]);

        while(q.size())
        {
            node p=q.top();
            if(d[i]>=p.t){q.pop();continue;}
            flag=1;
            printf("%d\n",p.x);
            break;
        }
        if(!flag)printf("-1\n");
    }
    return 0;
}
