#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>;

#define INF (1LL<<61)
#define MOD 1000000007LL 
//#define MOD 998244353LL
#define EPS (1e-10)

#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 ull = unsigned long long;
using umat = vector<vector<ull>>;

vector<mint> F, Finv;

void solve()
{
    char s[1];
    ll n, k;
    scanf("%1s(%ld,%ld)", s, &n, &k);
    if(s[0] == 'C') {
        if(n < k) PR(0);
        else PR((F[n]*Finv[n-k]*Finv[k]).x);
    }
    else if(s[0] == 'P') {
        if(n < k) PR(0);
        else PR((F[n]*Finv[n-k]).x);
    }
    else {
        if(n == 0 && k == 0) PR(1);
        else if(n == 0) PR(0);
        else PR((F[n+k-1]*Finv[n-1]*Finv[k]).x);
    }
}

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

    ll M = 2000000;
    F = vector<mint>(M+1, mint(1));
    Finv = vector<mint>(M+1, mint(1));
    REP(i,2,M+1) {
        F[i] = mint(i)*F[i-1];
        Finv[i] = F[i].inv();
    }

    REP(i,0,T) solve();

    return 0;
}

/*



*/