没学过 ntt 指路：https://blog.csdn.net/tenkuo/article/details/150500190

#include<bits/stdc++.h>
using namespace std;
 
typedef long long LL;
typedef __int128 i128;
const int N = 3e6 + 10;
const LL MOD = 1000000007;
 
const LL P1 = 998244353;
const LL P2 = 1004535809;
const LL P3 = 469762049;
 
LL a1[N], b1[N], a2[N], b2[N], a3[N], b3[N];
LL ans1[N], ans2[N], ans3[N];
LL temp_a[N], temp_b[N];
int limit, l, r[N];
int n, m;
 
LL q_pow(LL a, LL b, LL P) {
    LL c = 1;
    while (b) {
        if (b & 1) c = (i128)c * a % P;
        a = (i128)a * a % P;
        b >>= 1;
    }
    return c;
}
 
// NTT 实现
// type = 1: 正变换, type = -1: 逆变换
void ntt(LL *a, int type, LL P) {
    for (int i = 0; i < limit; i++) {
        if (i < r[i]) swap(a[i], a[r[i]]);
    }
    
    for (int mid = 1; mid < limit; mid <<= 1) {
        // 计算原根 g 的 (P-1)/(2*mid) 次方
        LL Wn = q_pow(3, (P - 1) / (mid << 1), P);
        if (type == -1) {
            Wn = q_pow(Wn, P - 2, P);  // 逆变换取逆
        }
        
        for (int R = (mid << 1), j = 0; j < limit; j += R) {
            LL w = 1;
            for (int k = 0; k < mid; k++, w = (i128)w * Wn % P) {
                LL x = a[j + k];
                LL y = (i128)a[j + mid + k] * w % P;
                a[j + k] = (x + y) % P;
                a[j + mid + k] = (x - y + P) % P;
            }
        }
    }
    
    // 如果是逆变换，乘以 inv_limit
    if (type == -1) {
        LL inv_limit = q_pow(limit, P - 2, P);
        for (int i = 0; i < limit; i++) {
            a[i] = (i128)a[i] * inv_limit % P;
        }
    }
}
 
void cvlt(LL *a, LL *b, LL *result, LL P) {
    // 复制数据并清零剩余部分
    for (int i = 0; i < limit; i++) {
        temp_a[i] = (i < n) ? a[i] % P : 0;
        temp_b[i] = (i < m) ? b[i] % P : 0;
    }
    
    // 正变换
    ntt(temp_a, 1, P);
    ntt(temp_b, 1, P);
    
    // 点乘
    for (int i = 0; i < limit; i++) {
        temp_a[i] = (i128)temp_a[i] * temp_b[i] % P;
    }
    
    // 逆变换
    ntt(temp_a, -1, P);
    
    // 复制结果
    for (int i = 0; i < limit; i++) {
        result[i] = temp_a[i];
    }
}
 
LL inv(LL a, LL mod) {
    return q_pow(a, mod - 2, mod);
}
 
LL crt(LL r1, LL r2, LL r3) {
    LL k1 = (r2 - r1) % P2;
    if (k1 < 0) k1 += P2;
    k1 = (i128)k1 * inv(P1 % P2, P2) % P2;
    
    i128 x12 = (i128)r1 + (i128)P1 * k1;
    i128 P12 = (i128)P1 * P2;
    
    LL r3_mod = (LL)(x12 % P3);
    LL k2 = (r3 - r3_mod) % P3;
    if (k2 < 0) k2 += P3;
    k2 = (i128)k2 * inv((LL)(P12 % P3), P3) % P3;
    
    i128 x = x12 + P12 * k2;
    x %= MOD;
    
    return (LL)x;
}
 
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    cin >> n >> m;
    
    for (int i = 0; i < n; i++) {
        cin >> a1[i];
        a2[i] = a3[i] = a1[i];
    }
    for (int i = 0; i < m; i++) {
        cin >> b1[i];
        b2[i] = b3[i] = b1[i];
    }
    
    limit = 1;
    l = 0;
    while (limit < n + m - 1) {
        limit <<= 1;
        l ++;
    }
    
    // 计算位逆序
    for (int i = 0; i < limit; i++) {
        r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
    }
    
    // 分别在三个模数下计算卷积
    cvlt(a1, b1, ans1, P1);
    cvlt(a2, b2, ans2, P2);
    cvlt(a3, b3, ans3, P3);
    
    // 输出结果
    for (int i = 0; i < n + m - 1; i++) {
        LL res = crt(ans1[i], ans2[i], ans3[i]);
        cout << res << " ";
    }
    cout << "\n";
    
    return 0;
}