三模数NTT WA 0pts求助

#include<bits/stdc++.h>
using namespace std;
#define N 114514
#define M 19198100
#define ll long long
#define re register int
ll n,m,p,a[M],b[M],r[M],ans[4][M];
//const ll A=998244353,B=469762049,C=1004535809,g=3; //三个原根为3的的1e9的模数
const ll P[3]={998244353,469762049,1004535809};
ll qpow(ll a,ll b,ll mod){
    ll ans=1;
    //cout<<"QWQ";
    while(b){
        if(b&1) ans=ans*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return ans;
}
//板子 
void NTT(ll n,ll *a,ll mod,ll g){
    for(ll i=0;i<n;++i)
        if(i<r[i]) swap(a[i],a[r[i]]); //位逆序置换
    for(ll i=2,h=1;i<=n;i<<=1,h<<=1){
        ll w=qpow(g,(mod-1)/i,mod);
        for(ll j=0;j<n;j+=i){
            ll W=1;
            for(ll k=0;k<h;++k,W=W*w%mod){
                ll x=a[j+k],y=W*a[h+j+k]%mod;
                a[j+k]=(x+y)%mod;
                a[h+j+k]=(x-y+mod)%mod;
            }
        }
    }
    //for(ll i=0;i<=n;++i) cout<<a[i]<<" ";
    //cout<<'\n';
}
ll x[M],y[M];
void work(ll n,ll *a,ll *b,ll mod,ll *ans){
    memcpy(x,a,n*sizeof(ll));
    memcpy(y,b,n*sizeof(ll));
    NTT(n,x,mod,3);
    NTT(n,y,mod,3);
    for(ll i=0;i<n;++i) x[i]=(ll)x[i]*y[i]%mod;//如果写的是int这里转成longlong一定要加
    NTT(n,x,mod,qpow(3,mod-2,mod));
    ll inv=qpow(n,mod-2,mod); //还是逆元
    for(ll i=0;i<n;++i) ans[i]=(ll)x[i]*inv%mod;
}
int main(){
    //三模数NTT 
    cin>>n>>m>>p;
    for(ll i=0;i<=n;++i) cin>>a[i];
    for(ll i=0;i<=m;++i) cin>>b[i];
    ll l=0,k=1;
    while(k<=n+m) k<<=1,l++;
    for(ll i=0;i<k;++i) r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));
    ll A=P[0],B=P[1],C=P[2];
    //根据CRT，通过三个序列得到确定的数 
    work(k,a,b,A,ans[0]);
    work(k,a,b,B,ans[1]);
    work(k,a,b,C,ans[2]);
    for(ll i=0;i<=n+m;++i){ //头晕柿子
        /*  EXCRT合并三个序列 
            Ai+k1*p1=Bi+k2*p2
            Ai+k1*p1≡Bi(mod p2)
            k1≡(Bi-Ai)/p1 (mod p2)
            得前两项解u=A1+k1*p1
        */
        ll u=ans[0][i]+((ans[1][i]-ans[0][i]+B)*qpow(A,B-2,B)%B)*A;
        /*
            u+k3*p1*p2=Ci+k4*p3
            u+k3*p1*p2≡Ci(mod p3)
            k3≡(Ci-u)/(p1*p2) (mod p3)
            得道三项通解v=u+k3*p1*p2
            ans=v%p 
        */
        ll v=(u%p+(ans[2][i]-u%C+C)*qpow(A*B,C-2,C)%C*A%p*B%p)%p;
        cout<<v<<" ";
    }
    return 0;
}
/*
5 8 28
19 32 0 182 99 95
77 54 15 3 98 66 21 20 38
*/