#pragma GCC optimize(2)
#include <bits/stdc++.h>
#define endl '\n'
#define int long long
#define vi vector<int>
using namespace std;
typedef pair<int, int> pii;
constexpr int N=210;
constexpr int inf = 1e18;
struct A
{
    int a;
    bool operator<(const A &t) const
    {
        return a < t.a;
    }
};
// priority_queue<pii, vector<pii>, greater<pii>> heap;
// cout << setprecision(n)
void solve()
{
    int n;
    cin>>n;
    // f[n][150][150][150]
    vector<array<int,3>> a(n+1);
    for(int i=1;i<=n;i++)
        for(int j=0;j<3;j++)
            cin >> a[i][j];
        
    vector f(3,vector(N,vector(N,vector(N,0ll))));
    for(int i=0;i<n;i++){
        //0
        int nxt=i&1^1,cur=i&1;
        for(int j=0;j<N;j++)
            for(int k=0;k<N;k++){
                int t=0;
                int tj = j + (j != 0), tk = k + (k != 0);
                if(tj<N&&tk<N){
                    for (int l = 0; l < N; l++)
                        t = max(t, f[cur][l][j][k]);
                        f[nxt][1][tj][tk] = max(f[nxt][1][tj][tk], t+a[i+1][0] - j - k);
                }
            
            }

        //1
        for(int j=0;j<N;j++)
            for(int k=0;k<N;k++){
                int t=0;
                int tj = j + (j != 0), tk = k + (k != 0);
                if (tj < N && tk < N)
                {
                    for (int l = 0; l < N; l++)
                        t = max(t, f[cur][j][l][k]);
                        f[nxt][tj][1][tk] = max(f[nxt][tj][1][tk], t+a[i+1][1] - j - k);
                }
            }

        //2
        for (int j = 0; j < N; j++)
            for (int k = 0; k < N; k++)
            {
                int t = 0;
                int tj = j + (j != 0), tk = k + (k != 0);
                if (tj < N && tk < N)
                {
                    for (int l = 0; l < N; l++)
                        t = max(t, f[cur][j][k][l]);
                        f[nxt][tj][tk][1] = max(f[nxt][tj][tk][1], t+a[i+1][2] - j - k);
                }
            }
    }

    int ans=0;
    for (int j = 0; j < N; j++)
        for (int k = 0; k < N; k++)
            ans=max({ans,(f[n&1][1][j][k]),(f[n&1][j][1][k]),(f[n&1][j][k][1])});
            
    // cout<<f[2][0][1][2]<<' '<<f[1][0][1][0]<<' '<<f[1][0][0][1]<<endl;
    cout << ans << endl;
}
signed main()
{
    // freopen("data3.in", "r", stdin);
    // freopen("data3.out", "w", stdout);
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t = 1;
    cin>>t;
    while (t--)
    {
        solve();
    }

    return 0;
}