Problem: Soon there will be held the world's largest programming contest, but the testing system still has m bugs. The contest organizer, a well-known university, has no choice but to attract university students to fix all the bugs. The university has n students able to perform such work. The students realize that they are the only hope of the organizers, so they don't want to work for free: the i-th student wants to get ci 'passes' in his subjects (regardless of the volume of his work).

Bugs, like students, are not the same: every bug is characterized by complexity aj, and every student has the level of his abilities bi. Student i can fix a bug j only if the level of his abilities is not less than the complexity of the bug: bi ≥ aj, and he does it in one day. Otherwise, the bug will have to be fixed by another student. Of course, no student can work on a few bugs in one day. All bugs are not dependent on each other, so they can be corrected in any order, and different students can work simultaneously.

The university wants to fix all the bugs as quickly as possible, but giving the students the total of not more than s passes. Determine which students to use for that and come up with the schedule of work saying which student should fix which bug.

Input

The first line contains three space-separated integers: n, m and s (1 ≤ n, m ≤ 105, 0 ≤ s ≤ 109) — the number of students, the number of bugs in the system and the maximum number of passes the university is ready to give the students.

The next line contains m space-separated integers a1, a2, ..., am (1 ≤ ai ≤ 109) — the bugs' complexities.

The next line contains n space-separated integers b1, b2, ..., bn (1 ≤ bi ≤ 109) — the levels of the students' abilities.

The next line contains n space-separated integers c1, c2, ..., cn (0 ≤ ci ≤ 109) — the numbers of the passes the students want to get for their help.

Output

If the university can't correct all bugs print "NO".

Otherwise, on the first line print "YES", and on the next line print m space-separated integers: the i-th of these numbers should equal the number of the student who corrects the i-th bug in the optimal answer. The bugs should be corrected as quickly as possible (you must spend the minimum number of days), and the total given passes mustn't exceed s. If there are multiple optimal answers, you can output any of them.

Examples

Input

3 4 9
1 3 1 2
2 1 3
4 3 6


Output

YES
2 3 2 3


Input

3 4 10
2 3 1 2
2 1 3
4 3 6


Output

YES
1 3 1 3


Input

3 4 9
2 3 1 2
2 1 3
4 3 6


Output

YES
3 3 2 3


Input

3 4 5
1 3 1 2
2 1 3
5 3 6


Output

NO

Note

Consider the first sample.

The third student (with level 3) must fix the 2nd and 4th bugs (complexities 3 and 2 correspondingly) and the second student (with level 1) must fix the 1st and 3rd bugs (their complexity also equals 1). Fixing each bug takes one day for each student, so it takes 2 days to fix all bugs (the students can work in parallel).

The second student wants 3 passes for his assistance, the third student wants 6 passes. It meets the university's capabilities as it is ready to give at most 9 passes.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
public final class preparing_for_contest
{
    static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	static FastScanner sc=new FastScanner(br);
    static PrintWriter out=new PrintWriter(System.out);
	static Random rnd=new Random();
	static int n,m;
	static long s;
	static Pair[] a;
	static Node[] b;
	static TreeMap<Integer,Integer> m1;
	static int[] res;
	static TreeMap<Integer,ArrayDeque<Integer>> ts;
	
	static void put(int curr,int val)
	{
		if(m1.get(curr)==null)	
		{
			m1.put(curr,val);
		}
		else
		{
			m1.put(curr,m1.get(curr)+val);
		}
		if(m1.get(curr)<=0)
		{
			m1.remove(curr);
		}
	}
	
	static boolean check(int mid)
	{
		int j=n-1;long total=0;m1=new TreeMap<>();
		for(int i=m-1;i>=0;)
		{
			while(j>=0 && b[j].l1>=a[i].val)
			{
				put(b[j].l2,1);j--;
			}
			if(true)
			{
				if(m1.size()>0)
				{
					Integer curr=m1.firstKey();
					if(curr!=null && total+curr<=s)
					{
						put(curr,-1);total+=curr;i-=mid;
					}
					else
					{
						return false;
					}
				}
				else
				{
					return false;
				}
			}
		}
		return true;
	}
	
	static void add(int idx,int idx2)
	{
		if(ts.get(idx)==null)
		{
			ArrayDeque<Integer> ad=new ArrayDeque<>();ad.add(idx2);ts.put(idx,ad);
		}
		else
		{
			ArrayDeque<Integer> ad=ts.get(idx);ad.add(idx2);ts.put(idx,ad);
		}
	}
	
	static Integer get()
	{
		Integer key=ts.firstKey();
		ArrayDeque<Integer> ad=ts.get(key);
		int now=ad.removeFirst();
		if(ad.size()==0)
		{
			ts.remove(key);
		}
		return now;
	}
	
	static void build(int mid)
	{
		int j=n-1;long total=0;ts=new TreeMap<>();int curr_idx=-1,cnt=-1;
		for(int i=m-1;i>=0;i--)
		{
			while(j>=0 && b[j].l1>=a[i].val)
			{
				add(b[j].l2,b[j].idx);j--;
			}
			if(cnt>0)
			{
				res[a[i].idx]=curr_idx;cnt--;
			}
			else
			{
				curr_idx=get();res[a[i].idx]=curr_idx;cnt=mid-1; 
			}
		}
	}
	
    public static void main(String args[]) throws Exception
    {
		n=sc.nextInt();m=sc.nextInt();s=sc.nextInt();a=new Pair[m];b=new Node[n];int low=1,high=m+1;
		for(int i=0;i<m;i++)
		{
			a[i]=new Pair(i,sc.nextInt());
		}
		for(int i=0;i<n;i++)
		{
			b[i]=new Node(i,sc.nextInt(),0);
		}
		for(int i=0;i<n;i++)
		{
			b[i].l2=sc.nextInt();
		}
		Arrays.sort(a);Arrays.sort(b);boolean ans=false;
		while(low<high)
		{
			int mid=(low+high)>>1;
			if(check(mid))
			{
				ans=true;high=mid;
			}
			else
			{
				low=mid+1;
			}
		}
		if(low<=m)
		{
			out.println("YES");res=new int[m];build(low);
			for(int i=0;i<m;i++)
			{
				out.print((res[i]+1)+" ");
			}
			out.println("");
		}
		else
		{
			out.println("NO");
		}
		out.close();
    }
}
class Node implements Comparable<Node>
{
	int idx,l1,l2;
	public Node(int idx,int l1,int l2)
	{
		this.idx=idx;this.l1=l1;this.l2=l2;
	}
	public int compareTo(Node x)
	{
		return Integer.compare(this.l1,x.l1);
	}
}
class Pair implements Comparable<Pair>
{
	int idx,val;
	public Pair(int idx,int val)
	{
		this.idx=idx;this.val=val;
	}
	public int compareTo(Pair x)
	{
		return Integer.compare(this.val,x.val);
	}
}
class FastScanner
{
    BufferedReader in;
    StringTokenizer st;

    public FastScanner(BufferedReader in) {
        this.in = in;
    }
	
    public String nextToken() throws Exception {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }
	
	public String next() throws Exception {
		return nextToken().toString();
	}
	
    public int nextInt() throws Exception {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() throws Exception {
        return Long.parseLong(nextToken());
    }

    public double nextDouble() throws Exception {
        return Double.parseDouble(nextToken());
    }
}