#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define int long long
#define debug(p) for (auto i : p)cerr << i << " "; cerr << endl;
#define debugs(p) for (auto i : p)cerr << i.first << " " << i.second << endl;
typedef pair<int, int> pll;
string yes = "YES";
string no = "NO";
constexpr int N = 1e6 + 7;
struct ACAM {
    int tr[N][26], cnt[N], ne[N], depth[N];//trie树存储，模式串叶子结点计数， 失配指针，字符串长度
    int tot = 0;
    int idx[N];//记录某个叶子节点对应的匹配串编号
    int from[N];//记录重复匹配串的映射节点
    int deg[N];//拓扑排序记录每个节点入度

    void ins(string s, int pos) //给定多个匹配串构建trie
    {
        int p = 0;
        for (int i = 0; i < s.size(); i++) {
            int u = s[i] - 'a';
            if (!tr[p][u])
                tr[p][u] = ++tot;  
            p = tr[p][u];
            cnt[p]++;
        }
        if(!idx[p])idx[p] = pos;
        from[pos] = idx[p];
    }
    void init(int n)
    {
        tot = 0;
        for (int i = 0; i <= n; i++)
        {
            for (int j = 0; j < 26; j++)tr[i][j] = 0;
            ne[i] = cnt[i] = idx[i] = depth[i] = from[i] = deg[i] = 0;
        }
    }
    void build() {
        queue<int>q;
        for (int i = 0; i < 26; i++) {
            if (tr[0][i])
            {
                q.push(tr[0][i]);
                depth[tr[0][i]] = 1;
            }
        }
        while (q.size()) {
            int p = q.front();
            q.pop();
            //枚举每个可能存在的子节点，如果存在就添加虚边并加入队列，不存在就维护节点
            for (int i = 0; i < 26; i++) {
                int s = tr[p][i];
                if (tr[p][i]) {
                    q.push(s);
                    depth[s] = depth[p] + 1;
                    ne[s] = tr[ne[p]][i];//子节点的虚边连向父节点虚边指向的另一个父节点的子节点
                    deg[ne[s]]++;
                }
                else
                    tr[p][i] = tr[ne[p]][i];
            }
        }
    }
}ac;
string ans;
void dfs(int p, int k)
{
    int tot = 0;
    if(k <= 0)return;
    for (int i = 0; i < 26; i++)
    {
        if(ac.tr[p][i])tot++;
    }
    if(k <= tot)return;
    int cnt = 0, sub = tot - 1;
    for (int i = 0; i < 26; i++)
    {
        if(!ac.tr[p][i])continue;
        cnt += ac.cnt[ac.tr[p][i]] - 1;
        
        if(tot + cnt >= k && ac.cnt[ac.tr[p][i]] >= 2)
        {
            if(k - sub >= 0)
            {
                ans.push_back('a' + i);
                p = ac.tr[p][i];
                // int mx = 0;
                // for (int j = 0; j < 26; j++)mx += ac.cnt[ac.tr[p][j]];
                if(ac.idx[p])sub++;
                dfs(p, k - sub);
            }
            return;
        }
        sub += ac.cnt[ac.tr[p][i]] - 1;
    }
}
int T;
void solve()
{
    int n, k;
    cin >> n >> k;
    vector<string>s(n + 1);
    for (int i = 1; i <= n; i++)
    {
        cin >> s[i];
        // if(T == 25000 - 3236)
        // {
        //     cout << s[i] << ";";
        // }
        ac.ins(s[i], i);
    }
    // if(T == 25000 - 3236)cout << k << ".";
    int p = 0;
    dfs(p, k);
    if(ans.empty())cout << "EMPTY" << endl;
    else cout << ans << endl;
    ac.init(ac.tot);
    ans.clear();
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    // int T = 1;
    cin >> T;
    while(T--)
    {
        solve();
    }
}