#pragma GCC optimization ("O3")

#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using vec = vector<ll>;
using mat = vector<vec>;
using pll = pair<ll,ll>;
using dvec = vector<double>;
using dmat = vector<dvec>;

#define INF (1LL<<61)
#define MOD 1000000007LL 
//#define MOD 998244353LL

#define PR(x) cout << (x) << endl
#define PS(x) cout << (x) << " "
#define REP(i,m,n) for(ll (i)=(m),(i_len)=(n);(i)<(i_len);++(i))
#define FORE(i,v) for(auto (i):v)
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((ll)(x).size())
#define REV(x) reverse(ALL((x)))
#define ASC(x) sort(ALL((x)))
#define DESC(x) {ASC((x)); REV((x));}
#define BIT(s,i) (((s)>>(i))&1)
#define pb push_back
#define fi first
#define se second

template<class T> inline int chmin(T& a, T b) {if(a>b) {a=b; return 1;} return 0;}
template<class T> inline int chmax(T& a, T b) {if(a<b) {a=b; return 1;} return 0;}

class mint {
public:
    ll x;
    mint(ll x=0) : x((x%MOD+MOD)%MOD) {}
    mint operator-() const {return mint(-x);}
    mint& operator+=(const mint& a) {if((x+=a.x)>=MOD) x-=MOD; return *this;}
    mint& operator-=(const mint& a) {if((x+=MOD-a.x)>=MOD) x-=MOD; return *this;}
    mint& operator*=(const mint& a) {(x*=a.x)%=MOD; return *this;}
    mint operator+(const mint& a) const {mint b(*this); return b+=a;}
    mint operator-(const mint& a) const {mint b(*this); return b-=a;}
    mint operator*(const mint& a) const {mint b(*this); return b*=a;}
    mint pow(ll t) const {if(!t) return 1; mint a=pow(t>>1); return (t&1?*this*a:a)*a;}
    mint inv() const {return pow(MOD-2);}
    mint& operator/=(const mint& a) {return *this*=a.inv();}
    mint operator/(const mint& a) const {mint b(*this); return b/=a;}
};
istream &operator>>(istream& is, mint& a) {ll t; is>>t; a=t; return is;}
ostream &operator<<(ostream& os, const mint& a) {return os<<a.x;}
using mvec = vector<mint>;
using mmat = vector<mvec>;

vec sieve(ll n)
{
    vec p(n+1);
    REP(i,0,n+1) p[i] = i;
    for(ll i=2; i*i<=n; ++i) {
        if(p[i] < i) continue;
        for(ll j=i*i; j<=n; j+=i) {
            if(p[j] == j) p[j] = i;
        }
    }
    return p;
}

map<ll,ll> factorize(ll n, vec& P)
{
    map<ll,ll> pe;
    ll p = P[n], e = 1;
    n /= p; 
    while(n > 1) {
        if(p == P[n]) ++e;
        else {
            pe[p%9] += e;
            p = P[n];
            e = 1;
        }
        n /= p;
    }
    if(p > 1) pe[p%9] += e;
    return pe;
}

ll modpow(ll a, ll n)
{
    if(n == 0) return 1;
    ll t = modpow(a, n/2);
    t = t*t%9;
    if(n%2) t = t*a%9;
    return t;
}

int main()
{
    ll T;
    cin >> T;

    ll M = 100010;
    vec P = sieve(M), I = {-1, 1, 5, -1, 7, 2, -1, 4, 8};
    mat F(M+1, vec(9));
    REP(i,2,M+1) {
        map<ll,ll> T = factorize(i, P);
        FORE(t,T) F[i][t.fi] += t.se;
    }
    REP(i,2,M+1) {
        REP(j,0,9) F[i][j] += F[i-1][j];
    }
    auto binom = [&](ll n, ll k) -> ll {
        ll x = 1, y = 1, z = 1;
        ll c = F[n][3]-F[k][3]-F[n-k][3];
        REP(i,2,9) {
            if(i%3 == 0) continue;
            x = x*modpow(i, F[n][i])%9;
            y = y*modpow(i, F[k][i])%9;
            z = z*modpow(i, F[n-k][i])%9;
        }
        if(c >= 2) return 0;
        return (c==1?3:1)*x*I[y]*I[z]%9;
    };

    while(T--) {
        string S;
        cin >> S;
        if(count(ALL(S), '0') == SZ(S)) {
            PR(0);
            continue;
        }
        ll ans = 0;
        REP(i,0,SZ(S)) ans = (ans+(S[i]-'0')*binom(SZ(S)-1, i))%9;
        PR(ans==0?9:ans);
    }

    return 0;
}

/*



*/