#include #include #include #include #include #include #include #define REP(i,s,n) for(int i=(int)(s);i<(int)(n);i++) using namespace std; typedef long long int ll; typedef vector VI; typedef vector VL; typedef pair PI; const ll mod = 1e9 + 7; /** * Segment Tree. This data structure is useful for fast folding on intervals of an array * whose elements are elements of monoid M. Note that constructing this tree requires the identity * element of M and the operation of M. * Header requirement: vector, algorithm * Verified by AtCoder ABC017-D (http://abc017.contest.atcoder.jp/submissions/660402) */ template class SegTree { int n; std::vector dat; BiOp op; I e; public: SegTree(int n_, BiOp op, I e) : op(op), e(e) { n = 1; while (n < n_) { n *= 2; } // n is a power of 2 dat.resize(2 * n); for (int i = 0; i < 2 * n - 1; i++) { dat[i] = e; } } /* ary[k] <- v */ void update(int k, I v) { k += n - 1; dat[k] = v; while (k > 0) { k = (k - 1) / 2; dat[k] = op(dat[2 * k + 1], dat[2 * k + 2]); } } void update_array(int k, int len, const I *vals) { for (int i = 0; i < len; ++i) { update(k + i, vals[i]); } } /* Updates all elements. O(n) */ void update_all(const I *vals, int len) { for (int k = 0; k < std::min(n, len); ++k) { dat[k + n - 1] = vals[k]; } for (int k = std::min(n, len); k < n; ++k) { dat[k + n - 1] = e; } for (int b = n / 2; b >= 1; b /= 2) { for (int k = 0; k < b; ++k) { dat[k + b - 1] = op(dat[k * 2 + b * 2 - 1], dat[k * 2 + b * 2]); } } } /* l,r are for simplicity */ I querySub(int a, int b, int k, int l, int r) const { // [a,b) and [l,r) intersects? if (r <= a || b <= l) return e; if (a <= l && r <= b) return dat[k]; I vl = querySub(a, b, 2 * k + 1, l, (l + r) / 2); I vr = querySub(a, b, 2 * k + 2, (l + r) / 2, r); return op(vl, vr); } /* [a, b] (note: inclusive) */ I query(int a, int b) const { return querySub(a, b + 1, 0, 0, n); } }; typedef vector VVL; VVL add(const VVL &a, const VVL &b) { assert (a.size() == b.size()); int n = a.size(); int m = a[0].size(); VVL ret(n, VL(m, 0)); REP(i, 0, n) { REP(j, 0, m) { ret[i][j] = (a[i][j] + b[i][j]) % mod; } } return ret; } VVL mul(const VVL &a, const VVL &b) { assert (a[0].size() == b.size()); int n = a.size(); int m = b.size(); int l = b[0].size(); VVL ret(n, VL(l, 0)); REP(i, 0, n) { REP(j, 0, m) { REP(k, 0, l) { ret[i][k] += a[i][j] * b[j][k]; ret[i][k] %= mod; } } } return ret; } vector elem_mul(const vector &abs1, const vector &abs2) { VVL a = mul(abs2[0], abs1[0]); VVL b = mul(abs1[1], abs2[1]); VVL s = add(abs1[2], mul(mul(abs1[1], abs2[2]), abs1[0])); return {a, b, s}; } VVL mat_s(int i) { VVL s(2, VL(3, 0)); s[0][0] = 6 * i; s[0][1] = 6 * i + 1; s[0][2] = 6 * i + 2; s[1][0] = 6 * i + 3; s[1][1] = 6 * i + 4; s[1][2] = 6 * i + 5; return s; } int main(void){ int n, q; cin >> n; VVL a(3, VL(1)), b(2, VL(1)); REP(i, 0, 3) { cin >> a[i][0]; } REP(i, 0, 2) { cin >> b[i][0]; } VVL unit3(3), unit2(2), dummy23(2, VL(3)); REP(i, 0, 3) { VL t(3); t[i] = 1; unit3[i] = t; } REP(i, 0, 2) { VL t(2); t[i] = 1; unit2[i] = t; } SegTree, vector (*) (const vector &, const vector &) > st(n + 1, elem_mul, vector{unit3, unit2, dummy23}); REP(i, 0, n + 1) { st.update(i, vector{unit3, unit2, mat_s(i)}); } cin >> q; REP(loop_cnt, 0, q) { string qty; int i; cin >> qty; if (qty == "a") { cin >> i; REP(k, 0, 3) { REP(j, 0, 3) { cin >> unit3[k][j]; } } vector cur = st.query(i, i); st.update(i, vector{unit3, cur[1], mat_s(i)}); } if (qty == "b") { cin >> i; REP(k, 0, 2) { REP(j, 0, 2) { cin >> unit2[k][j]; } } vector cur = st.query(i, i); st.update(i, vector{cur[0], unit2, mat_s(i)}); } if (qty == "ga") { cin >> i; VVL res = mul(st.query(0, i - 1)[0], a); cout << res[0][0] << " " << res[1][0] << " " << res[2][0] << endl; } if (qty == "gb") { cin >> i; vector tmp = st.query(i + 1, n); VVL res = add(mul(tmp[1], b), mul(mul(tmp[2], st.query(0, i)[0]), a)); cout << res[0][0] << " " << res[1][0] << endl; } } }