公式的话官方题解已经非常详细，这里就不再写公式了，大致推导为n步有x+y步是j,k两维移动，有n-x-y步是在i轴上移动。　在x+y的两维中，有y步是在y轴上移动,x步在x轴上移动。然后算上C(n,x+y)*C(x+y,x)*t1^(x/p)*t2(y/p)。就是每个点的贡献。这题卡常卡的太恶心了。

#include"bits/stdc++.h"
using namespace std;
typedef long long LL;
typedef long long ll;

const LL mod = 998244353;
LL pi = 2;
LL pe = 8388608, Pa[8388610];
LL per = pe-1;
LL fac[2][30],inv[2][30],mul0,mul1,mul2,qpe[1007];
inline LL qpow(LL a, LL b, LL m)
{
    LL ans = 1;
    a %= m;
    while(b > 0)
    {
        if(b&1) ans = ans*a%m;
        a = a*a%m;
        b >>= 1;
    }
    return ans;
}

inline LL fast_pow(LL a, LL b, LL m)
{
    LL ans = 1;
    LL mm = m-1;
    a &= mm;
    while(b > 0)
    {
        if(b&1) ans = ans*a&mm;
        a = a*a&mm;
        b >>= 1;
    }
    return ans;
}

inline LL exgcd(LL a,LL b,LL &x,LL &y) {
    if(b==0) {
        x=1,y=0;
        return a;
    }
    LL d=exgcd(b,a%b,y,x);
    y-=a/b*x;
    return d;
}
inline LL Inv(LL a,LL P){//求a在膜P下的逆
    if(a==0)return 1;
    LL x,y;
    exgcd(a,P,x,y);
    return (x%P+P)%P;
}


inline LL cal1(LL n) {
    LL ans=0;
    while(n > 0) {
        ans += (n>>1);
        n >>= 1;
    }
    return ans;
}

inline LL cal2(LL n) {
    if(n == 0) return 1;
    LL ans = 1;
    if(n/pe) ans = qpe[n/pe];
    ans = Pa[n&per];
    return cal2(n>>1)*ans&per;
}

inline LL cal(LL n,LL m, LL P) {
    LL cnt = cal1(n) - cal1(m) - cal1(n-m);
    LL a = cal2(n), b = cal2(m), c = cal2(n-m);
    LL ans =  ((a*Inv(b,pe)&per)*Inv(c,pe)&per)*fast_pow(pi,cnt,pe)&per;
    return ans;//范围可到达P*Pi，注意是否用快速乘
}


inline LL C(LL n, LL m, LL p, LL id)//组合数C(n, m) % p
{
    if(m > n) return 0;
    return fac[id][n] * inv[id][m]%p * inv[id][n - m] % p;
}
inline LL Lucas(LL n, LL m, LL p, LL id)
{
    if(m == 0) return 1;
    return C(n % p, m % p, p, id) * Lucas(n / p, m / p, p, id) % p;
}

inline LL solve(LL n,LL m,LL P) {
    if(m>n)return 0;
    LL ans = cal(n,m,P)*mul0%P;
    LL p = 7;
    LL t1 = Lucas(n,m,p,0)*mul1%P;
    p = 17;
    LL t2 = Lucas(n,m,p,1)*mul2%P;
    ans = (ans + t1 + t2)%P;
    return ans;
}

void init(LL P){
    mul0 = (P/pe)*Inv(P/pe,pe)%P;
    mul1 = (P/7)*Inv(P/7,7)%P;
    mul2 = (P/17)*Inv(P/17,17)%P;
    Pa[0] = Pa[1] = 1;
    for(LL j = 2; j <= pe; j++){
        if(j&1) Pa[j] = Pa[j-1]*j&per;
        else Pa[j] = Pa[j-1];
    }
    qpe[0] = Pa[pe];
    for(int i = 1; i <= 1000; i++)
        qpe[i] = qpe[i-1]*Pa[pe]%pe;
    LL p = 7;
    fac[0][0] = fac[0][1] = 1;
    inv[0][0] = inv[0][1] = 1;
    for(LL j = 2; j <= p; j++){
        fac[0][j] = fac[0][j-1]*j%p; 
        inv[0][j] = inv[0][j-1]*qpow(j,p-2,p)%p;
    }

    p = 17;
    fac[1][0] = fac[1][1] = 1;
    inv[1][0] = inv[1][1] = 1;
    for(LL j = 2; j <= p; j++){
        fac[1][j] = fac[1][j-1]*j%p; 
        inv[1][j] = inv[1][j-1]*qpow(j,p-2,p)%p;
    }
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
#endif
    int t1,t2,p,q,n,m;
    init(mod-1);
    while(~scanf("%d%d%d%d%d%d",&t1,&t2,&p,&q,&n,&m)){
        LL ans = 1;
        for(LL i = 1; i <= m; i++){
            int ax,ay,av;
            scanf("%d%d%d",&ax,&ay,&av);

            if(ax%p != 0 || ay%q != 0){
                continue;
            }
            LL x = ax/p , y = ay/q;
            if(x+y > n) continue;
            LL tmp = solve(x+y, x, mod-1)*solve(n,x+y,mod-1)%(mod-1)*qpow(t1,x,mod-1)%(mod-1)*qpow(t2,y,mod-1)%(mod-1);
            ans = ans*qpow(av,tmp,mod)%mod;
        }
        printf("%lld\n",(ans%mod+mod)%mod);
    }
    return 0;
}