#pragma clang diagnostic push
#pragma ide diagnostic ignored "cppcoreguidelines-narrowing-conversions"
#pragma ide diagnostic ignored "hicpp-signed-bitwise"
#pragma GCC optimize ("Ofast,unroll-loops")
#pragma GCC optimize("no-stack-protector,fast-math")

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<double> vd;
typedef vector<string> vs;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;
typedef vector<vll> vvll;
typedef vector<vvll> vvvll;
typedef vector<pii> vpii;
typedef vector<vpii> vvpii;
typedef vector<pll> vpll;
typedef vector<vpll> vvpll;
typedef vector<pdd> vpdd;
typedef vector<vd> vvd;
#define yn(ans) printf("%s\n", (ans)?"Yes":"No");
#define YN(ans) printf("%s\n", (ans)?"YES":"NO");
template<class T> bool chmax(T &a, T b) {
	if (a >= b) return false;
	a = b; return true;
}
template<class T> bool chmin(T &a, T b) {
	if (a <= b) return false;
	a = b; return true;
}
#define FOR(i, s, e, t) for ((i) = (s); (i) < (e); (i) += (t)) 
#define REP(i, e) for (int i = 0; i < (e); ++i) 
#define REP1(i, s, e) for (int i = (s); i < (e); ++i)
#define RREP(i, e) for (int i = (e); i >= 0; --i)
#define RREP1(i, e, s) for (int i = (e); i >= (s); --i)
#define all(v) v.begin(), v.end()
#define pb push_back
#define qb pop_back
#define pf push_front
#define qf pop_front
#define maxe max_element
#define mine min_element
ll inf = 1e18;
#define DEBUG printf("%d\n", __LINE__); fflush(stdout);
template<class T> void print(vector<T> &v, bool withSize = false) {
	if (withSize) cout << v.size() << endl;
	REP(i, v.size()) cout << v[i] << " "; 
	cout << endl;
}
mt19937_64 rng((unsigned int) chrono::steady_clock::now().time_since_epoch().count());

int __FAST_IO__ = []() {
	std::ios::sync_with_stdio(0);
	std::cin.tie(0);
	std::cout.tie(0);
	return 0;
}();

#define TESTS int t; cin >> t; while (t--)
#define TEST 
int main() {
    TEST {
        vi v(6);
        REP(i, 6) cin >> v[i];
        
        int mx = 1e6;
        vi dp(mx, 0);
        dp[0] = 1;
        REP(i, 6) {
            vi dp2 = dp;
            REP(j, mx + 1) {
                if (!dp[j]) continue;
                
                REP1(k, 1, 10) {
                    int base = 1;
                    int ret = 0;
                    REP(l, 6) {
                        ret += ((v[i] / base % 10) * k + (j / base % 10)) % 10 * base;
                        base *= 10;
                    }
                    dp2[ret] = 1;
                }
                
            }
            swap(dp, dp2);
        }
        int ans = 0;
        REP(i, mx) ans += dp[i];
        printf("%d\n", ans);
    }
    
    
    return 0;
}