Problem:
Score : 1800 points

Problem Statement

We have a square grid with N rows and M columns. Takahashi will write an integer in each of the squares, as follows:

First, write 0 in every square.

For each i=1,2,...,N, choose an integer k_i (0\leq k_i\leq M), and add 1 to each of the leftmost k_i squares in the i-th row.

For each j=1,2,...,M, choose an integer l_j (0\leq l_j\leq N), and add 1 to each of the topmost l_j squares in the j-th column.

Now we have a grid where each square contains 0, 1, or 2. Find the number of different grids that can be made this way, modulo 998244353.
We consider two grids different when there exists a square with different integers.

Constraints

1 \leq N,M \leq 5\times 10^5

N and M are integers.

Input

Input is given from Standard Input in the following format:

N M

Output

Print the number of different grids that can be made, modulo 998244353.

Sample Input 1

1 2

Sample Output 1

8

Let (a,b) denote the grid where the square to the left contains a and the square to the right contains b. Eight grids can be made: (0,0),(0,1),(1,0),(1,1),(1,2),(2,0),(2,1), and (2,2).

Sample Input 2

2 3

Sample Output 2

234

Sample Input 3

10 7

Sample Output 3

995651918

Sample Input 4

314159 265358

Sample Output 4

70273732

Solution in C++ (Clang 10.0.0) (judge verdict: Accepted)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define int long long
using namespace std;
const int maxn=500086;
const int mod=998244353;
int qpow(int a,int b,int p)
{
	int res=1,base=a%p;
	while(b) 
	{
		if(b&1) res=res*base%p;
		b>>=1,base=base*base%p;
	}
	return res;
}
int jc[maxn],inv[maxn];
int n,m;
int C(int n,int m)
{
	if(n<m) return 0;
	return jc[n]*inv[m]%mod*inv[n-m]%mod;
}
int F(int i)
{
	int fir=C(n,i)*C(m,i)%mod;
	int sec=qpow(m+1,n-i,mod)*qpow(n+1,m-i,mod)%mod;
	return fir*sec%mod*jc[i]%mod;
}
signed main()
{
	scanf("%lld%lld",&n,&m);
	jc[0]=1; for(int i=1;i<maxn;++i) jc[i]=jc[i-1]*i%mod;
	inv[0]=inv[1]=1; for(int i=2;i<maxn;++i) inv[i]=mod-(mod/i)*inv[mod%i]%mod;
	for(int i=2;i<maxn;++i) inv[i]=inv[i]*inv[i-1]%mod;
	int ans=0; 
	for(int i=0,type=1;i<=n&&i<=m;++i,type=mod-type) ans=(ans+type*F(i))%mod;
	printf("%lld\n",ans);
}