We all use cell phone today. And we must be familiar with the intelligent English input method on the cell phone. To be specific, the number buttons may correspond to some English letters respectively, as shown below:
　　2 : a, b, c 3 : d, e, f 4 : g, h, i 5 : j, k, l 6 : m, n, o
　　7 : p, q, r, s 8 : t, u, v 9 : w, x, y, z
　　When we want to input the word “wing”, we press the button 9, 4, 6, 4, then the input method will choose from an embedded dictionary, all words matching the input number sequence, such as “wing”, “whoi”, “zhog”. Here comes our question, given a dictionary, how many words in it match some input number sequences?

#include <cstdio>

#include <cmath>

#include <cstring>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <set>

#include <vector>

#include <sstream>

#include <queue>

#include <typeinfo>

#include <fstream>

#include <map>

#include <stack>

typedef long long ll;

using namespace std;

//freopen("D.in","r",stdin);

//freopen("D.out","w",stdout);

#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)

#define test freopen("test.txt","r",stdin)

const int maxn=;

#define mod 1000000009

#define eps 1e-7

const int inf=0x3f3f3f3f;

const ll infll = 0x3f3f3f3f3f3f3f3fLL;

inline ll read()

{

    ll x=,f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

//**************************************************************************************

//2 : a, b, c    3 : d, e, f    4 : g, h, i    5 : j, k, l    6 : m, n, o

//7 : p, q, r, s  8 : t, u, v    9 : w, x, y, z

map<int,int> HH;

string s[];

map<string,int> H;

int ans[];

int main()

{

    int k=;

    for(int i=;i<=;i++)

    {

        if(i==||i==||i==||i==||i==||i==||i==)

            k++;

        HH[i]=k;

    }

    int t=read();

    while(t--)

    {

        H.clear();

        memset(ans,,sizeof(ans));

        int n=read(),m=read();

        for(int i=;i<n;i++)

            cin>>s[i];

        string s1;

        for(int i=;i<m;i++)

        {

            cin>>s1;

            for(int j=;j<s1.size();j++)

                s1[j]=char(HH[(s1[j]-'a')]+'');

            H[s1]++;

        }

        for(int i=;i<n;i++)

            printf("%d\n",H[s[i]]);

    }

}