//SOLUTIE 2_0_3_3 OPTIMIZARE DUPA SFATURI MARIAN CALL 26-11-2020
//2021-11-04
#include <iostream>
using namespace std;
int main() {
int n, m, mt[501][501], k;
cin >> n >> m;
for(int i = 1; i <= n; i++){
for(int j = 1; j <= m; j++){
cin >> mt[i][j];
}
}
cin >> k;
int x, i;
for(int p = 1; p <= k; p++){
cin >> x;
if (x < mt[1][1] || x > mt[n][m]) {
cout << "0NU\n";
break;
} else { // mt[1][1] <= x && x <= mt[n][m] ...val x se incadreaza in interv de valori ale elem matricei
//int i, j;
if (x - mt[1][1] <= mt[n][m] - x) { // x e mai aproape de coltul din stg sus
i = 1;
//int left = 1, right = m;
while (mt[i][1] <= x && x <= mt[i][m]) {
int left = 1, right = m;
while (left < right) {
int mid = (left + right) / 2;
if (mt[i][mid] < x) {
left = mid + 1;
} else { // x <= mt[i][mid]
right = mid;
}
}
cout << "x= " << x << " ";
if (mt[i][left] == x) {
cout << "1DA\n";
break;
} else {
cout << "1NU\n";
break;
}
if (i < n) {
i++;
}
}
} else { // x - mt[1][1] > mt[n][m] - x ...adica x e mai aproape de coltul din dr jos
i = n;
while (mt[i][1] <= x && x <= mt[i][m]) {
int left = 1, right = m;
while (left < right) {
int mid = (left + right) / 2;
if (mt[i][mid] < x) {
left = mid + 1;
} else { // x <= mt[i][mid]
right = mid;
}
}
cout << "x= " << x << " ";
if (mt[i][left] == x) {
cout << "2DA\n";
break;
} else {
cout << "2NU\n";
break;
}
if (i > 1) {
i--;
}
}
}
}
}
return 0;
}
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