#pragma GCC optimization ("O3") #include using namespace std; using ll = long long; using vec = vector; using mat = vector; using pll = pair; #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 inline int chmin(T& a, T b) {if(a>b) {a=b; return 1;} return 0;} template inline int chmax(T& a, T b) {if(a=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< struct SegmentTree { using FX = function; int n; FX fx; const X ex; vector dat; SegmentTree(int n_, FX fx_, X ex_) : n(), fx(fx_), ex(ex_) { int x = 1; while(n_ > x) x *= 2; n = x; dat = vector(n*2, ex); } void set(int i, X x) { dat[i+n-1] = x; } void build() { for (int k = n-2; k>=0; k--) dat[k] = fx(dat[k*2+1], dat[k*2+2]); } void update(int i, X x) { i += n-1; dat[i] = x; while(i > 0) { i = (i-1)/2; dat[i] = fx(dat[i*2+1], dat[i*2+2]); } } X output_sub(int a, int b, int k, int l, int r) { if(r <= a || b <= l) return ex; else if(a <= l && r <= b) return dat[k]; else { X vl = output_sub(a, b, k*2+1, l, (l+r)/2); X vr = output_sub(a, b, k*2+2, (l+r)/2, r); return fx(vl, vr); } } X output(int a, int b) { return output_sub(a, b, 0, 0, n); } }; struct S { mat A, B, T; }; mat matadd(mat A, mat B) { ll N = SZ(A), M = SZ(A[0]); REP(i,0,N) { REP(j,0,M) A[i][j] += B[i][j], A[i][j] %= MOD; } return A; } mat matmul(mat A, mat B) { ll N = SZ(A), M = SZ(A[0]), K = SZ(B[0]); mat C(N, vec(K)); REP(i,0,N) { REP(j,0,K) { REP(k,0,M) C[i][j] += A[i][k]*B[k][j], C[i][j] %= MOD; } } return C; } int main() { ll N; cin >> N; mat a(3, vec(1)), b(2, vec(1)); REP(i,0,3) cin >> a[i][0]; REP(i,0,2) cin >> b[i][0]; S ex = {mat(3, vec(3)), mat(2, vec(2)), mat(2, vec(3))}; REP(j,0,3) ex.A[j][j] = 1; REP(j,0,2) ex.B[j][j] = 1; auto fx = [](S l, S r) -> S { S x; x.A = matmul(r.A, l.A); x.B = matmul(l.B, r.B); x.T = matadd(l.T, matmul(l.B, matmul(r.T, l.A))); return x; }; SegmentTree st(N+1, fx, ex); REP(i,0,N+1) { S x = {mat(3, vec(3)), mat(2, vec(2)), mat(2, vec(3))}; REP(j,0,3) x.A[j][j] = 1; REP(j,0,2) x.B[j][j] = 1; REP(j,0,2) { REP(k,0,3) x.T[j][k] = 6*i+3*j+k; } st.set(i, x); } st.build(); ll Q; cin >> Q; while(Q--) { string t; cin >> t; if(t == "a") { ll i; cin >> i; S x = st.output(i, i+1); REP(j,0,3) { REP(k,0,3) cin >> x.A[j][k]; } st.update(i, x); } else if(t == "b") { ll i; cin >> i; S x = st.output(i, i+1); REP(j,0,2) { REP(k,0,2) cin >> x.B[j][k]; } st.update(i, x); } else if(t == "ga") { ll i; cin >> i; S x = st.output(0, i); mat ans = matmul(x.A, a); cout << ans[0][0] << " " << ans[1][0] << " " << ans[2][0] << endl; } else if(t == "gb") { ll i; cin >> i; S x = st.output(0, i+1); S y = st.output(i+1, N+1); mat ans = matmul(y.B, b); mat tmp = matmul(y.T, matmul(x.A, a)); ans = matadd(ans, tmp); cout << ans[0][0] << " " << ans[1][0] << endl; } } return 0; } /* */