This is a template problem.

Here are two polynomials. Please output the multiplied polynomials.

Input format

In the first line, two integers nn and mm represent the degree of two polynomials respectively.

The second line, n+1n+1 integers, represents the coefficients of the terms from 00 to nn of the first polynomial.

The third line m+1m+1 integers represents the coefficients of the second polynomial from 00 to mm.

Output format

A line of n+m+1n+m+1 integers represents the coefficients of the terms from 00 to n+mn+m of the multiplied polynomials.

Example 1

input

1 2
1 2
1 2 1

output

1 4 5 2

explanation

(1+2x)⋅(1+2x+x2)=1+4x+5x2+2x3(1+2x)⋅(1+2x+x2)=1+4x+5x2+2x3.

Limitations and covenants

0 ≤ n,m ≤ 1050 ≤ n,m ≤ 105, ensure that the coefficient in the input is greater than or equal to 00 and less than or equal to 99.

Time limit: 1s1s

Space limit: 256MB

Shocked!

The reason for TLE in the morning is that there is no const in the definition of prime number and original root!

NTT's board questions

Just exchange the unit yuan for the original one

#include<cstdio>
#include<algorithm>
#include<cmath>
#define swap(x,y) x ^= y, y ^= x, x ^= y
#define LL long long 
using namespace std;
const int MAXN = 3 * 1e6 + 10;
inline int read(){
    int x=0,f=1;char ch=' ';
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+(ch^48),ch=getchar();
    return x*f;
}
int N, M, limit = 1, L;
const int P = 998244353, G = 3, Gi = 332748118; 
LL a[MAXN], b[MAXN];
int r[MAXN];
inline LL fastpow(LL a, LL k) {
    LL base = 1;
    while(k) {
        if(k & 1) base = (base * a ) % P;
        a = (a * a) % P;
        k >>= 1;
    }
    return base % P;
}
inline void NTT(LL *A, int type) {
    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) {    
        LL Wn = fastpow( type == 1 ? G : Gi , (P - 1) / (mid << 1));
        for(int j = 0; j < limit; j += (mid << 1)) {
            LL w = 1;
            for(int k = 0; k < mid; k++, w = (w * Wn) % P) {
                 int x = A[j + k], y = w * A[j + k + mid] % P;
                 A[j + k] = (x + y) % P,
                 A[j + k + mid] = (x - y + P) % P;
            }
        }
    }
}
int main() {
    #ifdef WIN32
    freopen("a.in", "r", stdin);
    #endif
    N = read(); M = read();
    for(int i = 0; i <= N; i++) a[i] = (read() + P) % P;
    for(int i = 0; i <= M; i++) b[i] = (read() + P) % P;
    while(limit <= N + M) limit <<= 1, L++;
    for(int i = 0; i < limit; i++)
        r[i] = (r[i >> 1] >> 1) | ((i & 1) << (L - 1));    
    NTT(a, 1);NTT(b, 1);    
    for(int i = 0; i < limit; i++) a[i] = (a[i] * b[i]) % P;
    NTT(a, -1);    
    LL inv = fastpow(limit, P - 2);
    for(int i = 0; i <= N + M; i++)
        printf("%d ", (a[i] * inv) % P);
    return 0;
}