#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#include <unordered_map>

typedef long long ll;
const ll mod = 1e7 + 19;
const ll P = 1e6;
const ll Q = 1e5 + 10;
const ll N = 1e7 + 10;
const ll T = 110;
ll n, q, t; 
ll sum, ans, add, mul = 1, res;
ll inv[mod]; 
ll opt[Q], id[Q], v[Q], a[T], b[T];
std::queue < ll > que;

struct hash_table {
    ll head[P], nxt[N], key[N], val[N], cnt;

    ll & operator [] (const ll &x) {
        ll tmp = x % P;
        if (!head[tmp]) que.push(tmp);
        for (ll i = head[tmp]; i; i = nxt[i]) 
            if (key[i] == x) return val[i];
        nxt[++cnt] = head[tmp], head[tmp] = cnt;
        key[cnt] = x, val[cnt] = res;
        return val[cnt];
    } 

    void clear () {
        while (!que.empty()) {
            ll x = que.front();
            head[x] = 0;
            que.pop();
        }
        cnt = 0;
    }
} h;

int main() {
    inv[1] = 1;
    for (ll i = 2; i < mod; i++) 
        inv[i] = inv[mod % i] * (mod - mod / i) % mod;
    std::cin >> n >> q;
    for (ll i = 1; i <= q; i++) {
        std::cin >> opt[i];
        switch (opt[i]) {
            case 1: std::cin >> id[i] >> v[i]; break;
            case 2: std::cin >> v[i]; break;
            case 3: std::cin >> v[i]; break;
            case 4: std::cin >> v[i]; break;
            case 5: std::cin >> id[i]; break;
        }
    }
    std::cin >> t;
    for (ll i = 1; i <= t; i++)
        std::cin >> a[i] >> b[i];
    for (ll i = 1; i <= t; i++)
        for (ll j = 1; j <= q; j++) {
            ll now = (a[i] + j * b[i]) % q + 1;
            switch (opt[now]) {
                case 1: 
                    sum = (sum - h[id[now]] * mul % mod - add) % mod;
                    h[id[now]] = (v[now] - add) * inv[(mul + mod) % mod] % mod;
                    sum = (sum + h[id[now]] * mul % mod + add) % mod;
                    break;
                case 2: 
                    add = (add + v[now]) % mod;
                    sum = (sum + v[now] * n % mod) % mod;
                    break;
                case 3:
                    if (v[now] == 0) sum = 0, mul = 1, add = 0, res = 0, h.clear();
                    else mul = mul * v[now] % mod, add = add * v[now] % mod, sum = sum * v[now] % mod;
                    break;
                case 4: 
                    add = 0, mul = 1, sum = n * v[now] % mod, res = v[now], h.clear();
                    break;
                case 5: 
                    ans = (ans + h[id[now]] * mul % mod + add) % mod;
                    break;
                case 6: 
                    ans = (ans + sum) % mod;
                    break;
            }
        }
    std::cout << (ans + mod) % mod << '\n';
    return 0;
}