#pragma GCC optimize(3, "Ofast", "inline")
#include<bits/stdc++.h>
#define endl '\n'
#define INF 0x3f3f3f3f3f3f3f3f
#define inf 0x3f3f3f3f
#define lowbit(x) ((x)&(-x))
using namespace std;
using ll = long long;
using pii = pair<int, int>;
const int N = 1e3 + 7;
const int mod = 998244353;
int f[2][3][210][210];
int a[N][3];
void solve(){
    int n;
    cin >> n;
    for(int i=1;i<=n;i++){
        cin >> a[i][0] >> a[i][1] >> a[i][2];
    }
    memset(f,0,sizeof f);
    int now = 0;
    for(int i=0;i<n;i++){
        for(int x = 0;x < 3; x++) {
            for (int j = 0; j < 210&&j<=i; j++) {
                for (int k = 0; k < 210&&k<=i; k++) {
                    int pj = j ? j + 1 : 0, pk = k ? k + 1 : 0;
                    if(pj<210&&pk<210)f[now][x][j][k] = max(f[now][x][pj][pk],f[now^1][x][j][k] - pj - pk + a[i + 1][x]);
                    if(pk<210)f[now][(x + 1) % 3][pk][1] = max(f[now][(x + 1) % 3][pk][1],f[now ^ 1][x][j][k] - 1 - pk + a[i + 1][(x + 1) % 3]);
                    if(pj<210)f[now][(x + 2) % 3][1][pj] = max(f[now][(x + 2) % 3][1][pj],f[now ^ 1][x][j][k] - 1 - pj + a[i + 1][(x + 2) % 3]);
                }
            }
        }
        now ^= 1;
    }
    int ans = 0;
    for(int x = 0; x < 3; x ++){
        for(int j = 0;j< 210;j ++){
            for(int k = 0;k < 210;k ++){
                ans = max(ans,f[now ^ 1][x][j][k]);
            }
        }
    }
    cout << ans << endl;
}
int main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
#ifdef qzhfx
    freopen("1.in", "r", stdin);
    freopen("1.out", "w", stdout);
#endif
    int t;
    t = 1;
    cin >> t;
    while(t--){
        solve();
    }
    return 0;
}