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/* -----------------------
Autor: Tomasz Boguslawski
-------------------------- */
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<fstream>
#include<iomanip>
#include<string>
#include<sstream>
#include<cstring>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include <fstream>
#include<math.h>

#define FOR(x, b, e) for(long x = b; x <= (e); x++)
#define FORD(x, b, e) for(long x = b; x >= (e); x--)
#define VAR(v, n) __typeof(n) v = (n)
#define ALL(c) (c).begin(), (c).end()
#define FOREACH(i, c) for(VAR(i, (c).begin()); i != (c).end(); ++i)
#define DEBUG if (debug)
#define MIN(a,b) ((a>b)?b:a)
#define MAX(a,b) ((a>b)?a:b)
#define LL long long

using namespace std;

long n;

class Node;

class Node
{
public:
    long num;
    long neiC;
    vector<long> nei;
    long col[2]; // col[0]-poczatkowy, col[1]-koncowy // 0-czerwone, 1-czarne
    void reset() { neiC=0; nei.clear(); }
    void addNeighbour(long num) { neiC++; nei.push_back(num);}
    bool isFunny();
};

Node nodes[510000];

long maxDegree()
{
    long md=0;
    FOR(i,1,n) if (nodes[i].neiC>md) md=nodes[i].neiC;
    return md;
}


/// mozliwe odpowiedzi:
/// 0 - nie zachodzi oczywisty przypadek
/// 1 - przypadek oczywisty, odpowiedz TAK
/// 2 - przypadek oczywisty, odpowiedz NIE
long obviousCase()
{
    long initially0=0; long initially1=0;
    long finally0=0; long finally1=0;
    FOR(i,1,n) if (nodes[i].col[0]==0) initially0++; else initially1++;
    FOR(i,1,n) if (nodes[i].col[1]==0) finally0++; else finally1++;
    if (initially0==n)
    {
        // wszystkie czerwone na poczatku
        if (finally0==n) return 1; // jest OK
        else return 2; // nie da sie
    }
    else if (initially1==n)
    {
        // wszystkie czarne na poczatku
        if (finally1==n) return 1; // jest OK
        else return 2; // nie da sie
    }
    return 0; // przypadek nie jest oczywisty
}

long anyLeaf()
{
    FOR(i,1,n) if (nodes[i].neiC==1) return i;
}

long countChanges(long leaf, long which)
{
    //cout << "Counting changes in " << which << " from leaf " << leaf << "\n";
    long cur=nodes[leaf].nei[0];
    long pop=leaf;
    long changes=0;
    if (nodes[leaf].col[which]!=nodes[cur].col[which]) changes=1;
    long next;
    while (nodes[cur].neiC>1)
    {
        next=nodes[cur].nei[0];
        if (next==pop) next=nodes[cur].nei[1];
        pop=cur; cur=next;
        if (nodes[pop].col[which]!=nodes[cur].col[which]) changes++;
    }
    //cout << " changes=" << changes << "\n";
    return changes;

}

bool okInSimpleTree()
{
    // maxDegree is 2
    long leaf=anyLeaf();
    long ch1=countChanges(leaf,0);
    long ch2=countChanges(leaf,1);
    if (ch1<ch2) return false;
    else if (ch1>ch2) return true;
    else return (nodes[leaf].col[0]==nodes[leaf].col[1]);
}

bool isBiparted()
{
    FOR(i,1,n) FOREACH (p,nodes[i].nei) if (nodes[i].col[1]==nodes[*p].col[1]) return false;
    return true;
}

bool areTheSame()
{
    FOR(i,1,n) if (nodes[i].col[0]!=nodes[i].col[1]) return false;
    return true;
}


/// MAIN
int main(int argc, char* argv[])
{
    // magic formula, which makes streams work faster:
	ios_base::sync_with_stdio(0);

    long t; cin >>t;
    string s1, s2;
    long a,b; bool ans;
    FOR(pt,1,t)
    {
        cin >> n;
        cin >> s1; cin >> s2;
        //cout << "s1=" << s1 << "#\n";
        //cout << "s2=" << s2 << "#\n";
        //cout << "Tree number " << t << " n=" << n << "\n";
        FOR(i,1,n) { nodes[i].col[0]=(s1[i-1]=='1'); nodes[i].col[1]=(s2[i-1]=='1'); nodes[i].reset(); nodes[i].num=i; }
        FOR(i,1,n-1)
        {
            cin >>a; cin >>b;
            nodes[a].addNeighbour(b); nodes[b].addNeighbour(a);
        }
        long oc=obviousCase();
        //cout << "oc=" << oc << "\n";
        if (oc==1) ans=true;
        else if (oc==2) ans=false;
        else
        {
            long md=maxDegree();
            //cout << "maxDegree=" << md << "\n";
            if (md<=2) ans=okInSimpleTree();
            else
            {
                bool isBi=isBiparted();
                if (isBi) { ans=areTheSame(); } //cout << "areTheSame=" << ans << "\n"; }
                else ans=true;
            }
        }
        if (ans) cout << "TAK\n"; else cout << "NIE\n";
    }
    return 0;
};