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#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <set>
#include <map>
#include <cmath>
#include <list>
#include <ctime>
#include <sstream>
#include <queue>
#include <stack>
#include <bitset>
#include <unordered_set>
#include <unordered_map>
using namespace std;
typedef vector<int> vi;
typedef pair<int,int> pii;
typedef long long ll;
typedef short int sint;
#define FOR(x, b, e) for(int x=(b); x<=(e); ++x)
#define FORD(x, b, e) for(int x=((int)(b))-1; x>=(e); --x)
#define REP(x, n) for(int x=0; x<(n); ++x)
#define ALL(c) c.begin(),c.end()
#define SIZE(x) ((int)((x).size()))
#define PB push_back
#define ST first
#define ND second
#define mp(x,y) make_pair(x,y)
#define DEBUG 1
#define debug(x) {if (DEBUG)cerr <<#x <<" = " <<x <<endl; }
#define debugv(x) {if (DEBUG) {cerr <<#x <<" = "; FOREACH(it, (x)) cerr <<*it <<", "; cout <<endl; }}
#define REMAX(a,b) (a)=max((a),(b));
#define REMIN(a,b) (a)=min((a),(b));
#define wez(n) int (n); scanf("%d",&(n));
#define wez2(n,m) int (n),(m); scanf("%d %d",&(n),&(m));

const int N = 1010;
set<int> tak[N], takIn[N], nie[N], nieIn[N];
int par[N];
int n, m;

bool isFree(int elem) {
	return SIZE(tak[elem]) == 0 && SIZE(nieIn[elem]) == 0;
}

void visit(int elem, set<int>& wid, set<int>& cur) {
	cur.insert(elem);
	wid.insert(elem);
	for (auto syn : tak[elem]) {
		if (wid.count(syn) == 0) {
			visit(syn, wid, cur);
		}
	}
	for (auto syn : takIn[elem]) {
		if (wid.count(syn) == 0) {
			visit(syn, wid, cur);
		}
	}
}

void printSet(set<int> elems) {
	for (auto elem : elems) printf("%d ", elem); printf("\n");
}

bool solve(set<int> elems, int parent) {
	// printSet(elems); printf(" with parent: %d\n", parent);
	// potencjalni ojcowie
	// znajdz ziomkow, ktorzy "tak" nie istnieje w zbiorze elems oraz brak "nie" -> na nich
	queue<int> Q;
	for (auto elem : elems) {
		if (isFree(elem)) {
			Q.push(elem);
		}
	}

	if (Q.empty()) {
		return false;
	}

	while (!Q.empty()) {
		int x = Q.front(); Q.pop();
		par[x] = parent;
		parent = x;
		// printf("sciagam: %d, nowy parent: %d\n", x, parent);
		elems.erase(x);
		for (auto bossWant : takIn[x]) {
			tak[bossWant].erase(x);
			if (isFree(bossWant)) {
				Q.push(bossWant);
			}
		}
	}

	// zostaly mi elementy ze nie ma jasnego szefa
	// podziel graf na spojne uzywajac krawedzi TAK, z obu tak oraz takIn
	set<int> wid, cur;
	for (auto x : elems) {
		if (wid.count(x) == 0) {
			// nowa spojna

			visit(x, wid, cur);
			// printf("graf indukowany: "); printSet(cur);
			for (auto elem : cur) {
				for (auto it = nie[elem].begin(); it != nie[elem].end(); ) {
					if (cur.count(*it) == 0) {
						it = nie[elem].erase(it);
					} else {
						++it;
					}
				}
				for (auto it = nieIn[elem].begin(); it != nieIn[elem].end(); ) {
					if (cur.count(*it) == 0) {
						it = nieIn[elem].erase(it);
					} else {
						++it;
					}
				}
			}
			bool isSol = solve(cur, parent);
			if (!isSol) {
				return false;
			}
			// printf("jus wdidziane: "); printSet(wid);
			cur.clear();
		}
	}
	return true;
}

int main() {
	ios_base::sync_with_stdio(false);
	cin >> n >> m;
	REP(i, m) {
		int a, b; char t;
		cin >> a >> b >> t;
		if (t == 'T') {
			tak[a].insert(b);
			takIn[b].insert(a);
		} else {
			nie[a].insert(b);
			nieIn[b].insert(a);
		}
	}
	set<int> elems;
	FOR(i, 1, n) {
		elems.insert(i);
	}
	bool result = solve(elems, 0);
	if (!result) {
		printf("NIE\n");
	} else {
		// printf("TAK\n");
		FOR(i, 1, n) printf("%d\n", par[i]);
	}
	return 0;
}