先考虑每一个洞穴，为了要打败所有的怪物，考虑每个怪兽要求的起始等级，然后求最大值。接下来考虑所有的洞穴，每个洞穴的起始等级要求不一样，可以先从小到大排序，因为对于给定等级，判定能否通关只需要一次线性扫描，所以可以二分答案。除了二分解法，其实也可以按照单个洞穴的思路来考虑所有的洞穴。

// Date: Sun Dec  3 10:23:28 2023

#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>

#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

using namespace std;

const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
const int dir[8][2] = {
    {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;

const ull Pr = 131;

#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])

#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nonempty(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())

#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z)                                                        \
  cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#else
#define debug1
#define debug2
#define debug3
#endif

const int N = 100010;
int t, n;
pair<ll, ll> b[N];
VI a[N];

bool check(ll x) {
  For(i, 0, n) {
    if (x < b[i].f1)
      return false;
    x += b[i].f2;
  }
  return true;
}

int main(void) {
#ifdef _DEBUG
  freopen("1561c.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  cin >> t;

  while (t--) {
    cin >> n;

    For(i, 0, n) {
      int len;
      cin >> len;
      a[i].clear();

      For(j, 0, len) {
        int x;
        cin >> x;
        a[i].pb(x);
      }

      ll ma = a[i][0];
      For(j, 1, len) { ma = max(ma, ll(a[i][j] - j)); }
      b[i].f1 = ma;
      b[i].f2 = len;
    }

    sort(b, b + n);

    ll l = 1, r = 1e18, mid;
    while (l < r) {
      mid = (l + r) / 2;
      if (check(mid))
        r = mid;
      else
        l = mid + 1;
    }

    cout << r + 1 << '\n';
  }

  return 0;
}