#include <bits/stdc++.h>
using namespace std;

ifstream fin("disjoint.in");
ofstream fout("disjoint.out");

const int MAX_SIZE = 100000;

int tree[MAX_SIZE + 1];
//int treeHeight[MAX_SIZE + 1];

int findRoot(int startNode) {
    int root = startNode;
    while (tree[root] != 0) {
        root = tree[root];
    }
    int currentNode = startNode;
    while (tree[currentNode] != 0) {
        int aux = currentNode;
        currentNode = tree[currentNode];
        tree[aux] = root;
    }
    return root;
}

void uniteTrees(int root1, int root2) {
   /* if (treeHeight[root2] > treeHeight[root1]) {
        tree[root2] = root1;
    } else {
        tree[root1] = root2;
    }
    if (treeHeight[root1] == treeHeight[root2]) {
        ++treeHeight[root2];
    } */
    tree[root2] = root1;
}

int main() {
    int noNodes, noOperations;
    fin >> noNodes >> noOperations;
    for (int i = 1; i <= noNodes; ++i) {
        treeHeight[i] = 1;
    }
    for (int i = 1; i <= noOperations; ++i) {
        int operation, node1, node2;
        fin >> operation >> node1 >> node2;
        if (operation == 1) {
            uniteTrees(findRoot(node1), findRoot(node2));
        } else {
            if (findRoot(node1) == findRoot(node2)) {
                fout << "DA\n";
            } else {
                fout << "NU\n";
            }
        }
    }
    return 0;
}
/*

1) Cazul din problema:

4 3
1 4 1
2 4 2
2 4 1
=>
NU
DA

2) Cazul cand toate nodurile formeaza o singura multime:

3 5
1 1 2
1 1 3
2 1 2
2 1 3
2 3 2
=>
DA
DA
DA

3) Cazul cand nu exista nici o operatie de tip 1 si atunci pentru orice operatie de tip 2 raspunsul va fi "NU":

3 3
2 2 1
2 3 1
2 3 2
=>
NU
NU
NU

4) Cazul in care nu exista nici o operatie de tip 2:

2 1
1 1 2
=>
(nimic)

5) Cazul cand inainte de operatiile de tipul 2 avem numai 2 submultimi:

4 6
1 1 2
1 2 3
1 4 5
1 4 6
2 1 4
2 5 6
=>
NU
DA

6) Cazul din problema:

4 6
1 1 2
1 3 4
2 1 3
2 1 2
1 1 3
2 1 4
=>
NU
DA
DA

*/