Problem: You are given n patterns p_1, p_2, ..., p_n and m strings s_1, s_2, ..., s_m. Each pattern p_i consists of k characters that are either lowercase Latin letters or wildcard characters (denoted by underscores). All patterns are pairwise distinct. Each string s_j consists of k lowercase Latin letters.

A string a matches a pattern b if for each i from 1 to k either b_i is a wildcard character or b_i=a_i.

You are asked to rearrange the patterns in such a way that the first pattern the j-th string matches is p[mt_j]. You are allowed to leave the order of the patterns unchanged.

Can you perform such a rearrangement? If you can, then print any valid order.

Input

The first line contains three integers n, m and k (1 ≤ n, m ≤ 10^5, 1 ≤ k ≤ 4) — the number of patterns, the number of strings and the length of each pattern and string.

Each of the next n lines contains a pattern — k characters that are either lowercase Latin letters or underscores. All patterns are pairwise distinct.

Each of the next m lines contains a string — k lowercase Latin letters, and an integer mt (1 ≤ mt ≤ n) — the index of the first pattern the corresponding string should match.

Output

Print "NO" if there is no way to rearrange the patterns in such a way that the first pattern that the j-th string matches is p[mt_j].

Otherwise, print "YES" in the first line. The second line should contain n distinct integers from 1 to n — the order of the patterns. If there are multiple answers, print any of them.

Examples

Input


5 3 4
_b_d
__b_
aaaa
ab__
_bcd
abcd 4
abba 2
dbcd 5


Output


YES
3 2 4 5 1 


Input


1 1 3
__c
cba 1


Output


NO


Input


2 2 2
a_
_b
ab 1
ab 2


Output


NO

Note

The order of patterns after the rearrangement in the first example is the following: 

  * aaaa
  * __b_
  * ab__
  * _bcd
  * _b_d



Thus, the first string matches patterns ab__, _bcd, _b_d in that order, the first of them is ab__, that is indeed p[4]. The second string matches __b_ and ab__, the first of them is __b_, that is p[2]. The last string matches _bcd and _b_d, the first of them is _bcd, that is p[5].

The answer to that test is not unique, other valid orders also exist.

In the second example cba doesn't match __c, thus, no valid order exists.

In the third example the order (a_, _b) makes both strings match pattern 1 first and the order (_b, a_) makes both strings match pattern 2 first. Thus, there is no order that produces the result 1 and 2.

Solution (in JAVA):

Solution:
//stan hu tao
//join nct ridin by first year culture reps
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import static java.lang.System.out;
import java.util.*;
import java.io.*;
import java.math.*;

public class x1476E
{
    static ArrayDeque<Integer>[] edges;
    public static void main(String hi[]) throws Exception
    {
        BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(infile.readLine());
        int N = Integer.parseInt(st.nextToken());
        int M = Integer.parseInt(st.nextToken());
        int K = Integer.parseInt(st.nextToken());
        String[] arr = new String[N];
        for(int i=0; i < N; i++)
            arr[i] = infile.readLine();
        String[] brr = new String[M];
        int[] match = new int[M];
        HashMap<String, Integer> check = new HashMap<String, Integer>();
        for(int i=0; i < M; i++)
        {
            st = new StringTokenizer(infile.readLine());
            brr[i] = st.nextToken();
            match[i] = Integer.parseInt(st.nextToken())-1;
            if(check.containsKey(brr[i]) && check.get(brr[i]) != match[i])
            {
                System.out.println("NO");
                return;
            }
            check.put(brr[i], match[i]);
            if(!valid(arr[match[i]], brr[i]))
            {
                System.out.println("NO");
                return;
            }
        }
        edges = new ArrayDeque[N];
        for(int i=0; i < N; i++)
            edges[i] = new ArrayDeque<Integer>();
        HashMap<String, Integer> map = new HashMap<String, Integer>();
        for(int i=0; i < N; i++)
            map.put(arr[i], i);
        for(int i=0; i < M; i++)
        {
            //create edges
            String base = brr[i];
            outer:for(int mask=0; mask < 1<<K; mask++)
            {
                String temp = "";
                for(int b=0; b < K; b++)
                {
                    if((mask&(1<<b)) == 0)
                        temp += base.charAt(b);
                    else
                        temp += '_';
                }
                if(map.containsKey(temp) && match[i] != map.get(temp))
                    edges[match[i]].add(map.get(temp));
            }
        }
        seen = new boolean[N];
        ArrayList<Integer> topsort = new ArrayList<Integer>();
        for(int i=0; i < N; i++)
            if(!seen[i])
                dfs(i, topsort);
        Collections.reverse(topsort);
        int[] pos = new int[N];
        for(int i=0; i < N; i++)
            pos[topsort.get(i)] = i;
        for(int v=0; v < N; v++)
            for(int next: edges[v])
                if(pos[v] > pos[next])
                {
                    System.out.println("NO");
                    return;
                }
        StringBuilder sb = new StringBuilder("YES\n");
        for(int x: topsort)
            sb.append((x+1)+" ");
        System.out.println(sb);
    }
    static boolean[] seen;
    public static void dfs(int curr, ArrayList<Integer> ls)
    {
        seen[curr] = true;
        for(int next: edges[curr])
            if(!seen[next])
                dfs(next, ls);
        ls.add(curr);
    }
    public static boolean valid(String pat, String word)
    {
        for(int i=0; i < pat.length(); i++)
            if(pat.charAt(i) != '_' && pat.charAt(i) != word.charAt(i))
                return false;
        return true;
    }
}