#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <string>
#include <iostream>
#include <map>
#include <cstdlib>
#include <list>
#include <set>
#include <queue>
#include <stack>using namespace std;typedef long long LL;
const int maxn = 205;
const int INF = INT_MAX / 3;struct Edge {int u,v,cap;Edge(int u,int v,int cap):u(u),v(v),cap(cap) {}
};int n,m,incnt[maxn],outcnt[maxn];
int deg[maxn],s,t;
vector<Edge> edges;
vector<int> e[maxn];void adde(int u,int v,int w) {int m = edges.size();edges.push_back(Edge(u,v,w));edges.push_back(Edge(v,u,0));e[u].push_back(m);e[v].push_back(m ^ 1);
}int level[maxn],q[maxn * 2],qs,qe;
bool bfs() {//建立层次网络memset(level,0,sizeof(level));level[s] = 1;qs = qe = 0;q[qe++] = s;while(qs < qe) {int now = q[qs++],nm = e[now].size();if(now == t) break;for(int i = 0;i < nm;i++) {Edge &ne = edges[e[now][i]];if(ne.cap && level[ne.v] == 0) {level[ne.v] = level[now] + 1;q[qe++] = ne.v;}}}return level[t];
}int dfs(int now,int alpha) {if(now == t) return alpha;int sum = 0,nm = e[now].size();for(int i = 0;i < nm;i++) {Edge &ne = edges[e[now][i]];if(level[now] + 1 == level[ne.v] && ne.cap && alpha) {int ret = dfs(ne.v,min(alpha,ne.cap));ne.cap -= ret; edges[e[now][i] ^ 1].cap += ret;sum += ret; alpha -= ret;}}if(sum == 0) level[now] = -1;return sum;
}void dinic() {while(bfs()) dfs(s,INF);
}bool solve() {s = 0; t = n + 1;//判断入度出度之差是否为偶数for(int i = 1;i <= n;i++) {deg[i] = incnt[i] - outcnt[i];if(deg[i] & 1) return false;}//建立容量网络for(int i = 1;i <= n;i++) {//如果入度小于出度,建立从起点到这个点的边，容量为deg/2if(deg[i] < 0) adde(s,i,-deg[i] / 2);//如果出度大于入读，建立从当前点到汇点的边，容量同样为deg/2if(deg[i] > 0) adde(i,t,deg[i] / 2);}//计算最大流dinic();//判断从源点出发的所有边是否满流int m = e[s].size();for(int i = 0;i < m;i++) {if(edges[e[s][i]].cap != 0) return false;}return true;
}int main() {int T; scanf("%d",&T);while(T--) {scanf("%d%d",&n,&m);edges.clear();for(int i = 0;i <= n + 1;i++) e[i].clear();memset(incnt,0,sizeof(incnt));memset(outcnt,0,sizeof(outcnt));for(int i = 1;i <= m;i++) {int u,v,c; scanf("%d%d%d",&u,&v,&c);//先将无向边全部作为有向边处理incnt[v]++; outcnt[u]++;//无向边存起来if(c == 0) adde(u,v,1);}if(solve()) puts("possible");else puts("impossible");}return 0;
}