#include <cstdio>#include <cstring>#include <iostream>#include <map>#include <set>#include <vector>#include <string>#include <queue>#include <deque>#include <bitset>#include <list>#include <cstdlib>#include <climits>#include <cmath>#include <ctime>#include <algorithm>#include <stack>#include <sstream>#include <numeric>#include <fstream>#include <functional>using namespace std;#define MP make_pair#define PB push_backtypedef long long LL;typedef unsigned long long ULL;typedef vector<int> VI;typedef pair<int,int> pii;const int INF = INT_MAX / 3;const double eps = 1e-8;const LL LINF = 1e17;const double DINF = 1e60;const int maxn = 1005;VI e[maxn];int n, m, vis[maxn][2], note[maxn][2];int dfs(int i, int j, int f) {    if(vis[i][j]) return note[i][j];    vis[i][j] = 1;    int &ret = note[i][j] = INF;    //如果当前节点不放灯（父亲节点必须放灯）    //总数为每一个节点的和 × 2000 + 1    //需要对根节点进行特判    int sum = f == -1 ? 0 : 1, mm = e[i].size();    if(j) {        for(int nx = 0;nx < mm;nx++) if(e[i][nx] != f) {            sum += dfs(e[i][nx], 0, i);        }        ret = sum;    }    //当前节点放灯    //总数为每一个节点的和*2000，然后如果父亲节点没有灯，再+1    sum = j == 0 ? 1 : 0;    for(int nx = 0;nx < mm;nx++) if(e[i][nx] != f) {        sum += dfs(e[i][nx], 1, i);    }    sum += 2000;    ret = min(sum,ret);    return ret;}int main() {    int T; scanf("%d",&T);    while(T--) {        memset(vis,0,sizeof(vis));        memset(note,0,sizeof(note));        scanf("%d%d",&n,&m);        for(int i = 0;i < n;i++) e[i].clear();        for(int i = 0;i < m;i++) {            int a,b; scanf("%d%d",&a,&b);            e[a].PB(b); e[b].PB(a);        }        //当前节点，父节点有没有点亮，父节点编号        //dfs(i,j,k)        //枚举森林中的每一棵树        int ans = 0;        for(int i = 0;i < n;i++) if(!vis[i][0]) {            ans += dfs(i,1,-1);        }        printf("%d %d %d\n",ans / 2000,m - ans % 2000,ans % 2000);    }    return 0;}