#include <iostream>
#include <string>
#include <vector>
using namespace std;

#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)

constexpr int MD = 101;
constexpr int MD6 = MD * 6;

char solve() {
    int N, K;
    cin >> N >> K;

    vector mat(3, vector<long long>(3));
    vector<int> A(N);
    string S;

    for (auto &vec : mat) {
        for (auto &x : vec) cin >> x;
    }

    for (auto &a : A) cin >> a;

    cin >> S;

    long long det = 0;
    REP(i, 3) {
        det += mat[0][i] * mat[1][(i + 1) % 3] * mat[2][(i + 2) % 3];
        det -= mat[0][i] * mat[1][(i + 2) % 3] * mat[2][(i + 1) % 3];
    }

    if (det == 0) return 'D';

    det = (max(det, -det) % MD6 + MD6) % MD6;

    string dp(MD6, 'D');
    REP(m, dp.size()) {
        if (m % 6) continue;
        dp.at(m) = ((m / 6) % MD >= K) ? 'K' : 'P';
    }

    IREP(i, A.size()) {
        const char c = S.at(i);
        auto dpprv = dp;
        REP(m, dp.size()) {
            const int nxtm = m * (A.at(i) + 1) % MD6;
            if (dp.at(nxtm) == c) dpprv.at(m) = c;
            if (dp.at(nxtm) == 'D' and dpprv.at(m) != c) dpprv.at(m) = 'D';
        }
        dp = dpprv;
    }

    return dp.at(det);
}

int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    int T;
    cin >> T;
    while (T--) cout << solve() << '\n';
}