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

typedef long long LL;

template < typename Tp >
void read(Tp &x) {
    x = 0; int fh = 1; char ch = 1;
    while(ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
    if(ch == '-') fh = -1, ch = getchar();
    while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    x *= fh;
}

const int MAXN = 50000 + 7;
const int MAXM = 10 + 3;

int n, m;

struct Info{
    int a[MAXM];
    Info(){
        a[0] = a[1] = a[2] = a[3] = a[4] = a[5] = a[6] = a[7] = a[8] = a[9] = a[10] = a[11] = a[12] = 0;
    }
    bool operator < (const Info &p) const {
        for(int i = 1; i < m; i++) {
            if(a[i] != p.a[i]) return a[i] < p.a[i];
        }
        return a[m] < p.a[m];
    }
}p[MAXN];

vector <Info> vec;

void Init(void) {
    read(n); read(m);
    for(int i = 1; i <= n; i++) {
        Info rev;
        for(int j = 1; j <= m; j++) {
            read(p[i].a[j]);
            rev.a[p[i].a[j]] = j;
        }
        vec.push_back(rev);
    }
}

int getans(Info ord, Info to) {
    for(int i = 1; i <= m; i++) {
        if(ord.a[i] != to.a[i]) return i - 1;
    }
    return m;
}

void Work(void) {
    sort(vec.begin(), vec.end());
    for(int i = 1; i <= n; i++) {
        auto it = lower_bound(vec.begin(), vec.end(), p[i]);
        int ans = 0;
        if(it == vec.end()) {
            --it;
            ans = getans(p[i], *it);
        }
        else {
            ans = getans(p[i], *it);
            --it;
            if(it != vec.end()) {
                ans = max(ans, getans(p[i], *it));
            }
        }
        printf("%d%c", ans, " \n"[i == n]);
    }
}

signed main(void) {
    int T; read(T);
    while(T--) {
        Init();
        Work();
    }
    return 0;
}

代码时间复杂度应该是 $O(nm \log n)$，但 TLE #2，不是很懂。