Problem:
Score : 2000 points

Problem Statement

You are participating in a quiz with N + M questions and Yes/No answers.

It's known in advance that there are N questions with answer Yes and M questions with answer No, but the questions are given to you in random order.

You have no idea about correct answers to any of the questions.
You answer questions one by one, and for each question you answer, you get to know the correct answer immediately after answering.

Suppose you follow a strategy maximizing the expected number of correct answers you give.

Let this expected number be P/Q, an irreducible fraction. Let M = 998244353.
It can be proven that a unique integer R between 0 and M - 1 exists such that P = Q \times R modulo M,
and it is equal to P \times Q^{-1} modulo M, where Q^{-1} is the modular inverse of Q.
Find R.

Constraints

1 \leq N, M \leq 500,000

Both N and M are integers.

Partial Score

1500 points will be awarded for passing the testset satisfying N = M and 1 \leq N, M \leq 10^5.

Input

Input is given from Standard Input in the following format:

N M

Output

Let P/Q be the expected number of correct answers you give if you follow an optimal strategy, represented as an irreducible fraction.
Print P \times Q^{-1} modulo 998244353.

Sample Input 1

1 1

Sample Output 1

499122178

There are two questions.
You may answer randomly to the first question, and you'll succeed with 50% probability.
Then, since you know the second answer is different from the first one, you'll succeed with 100% probability.

The expected number of your correct answers is 3 / 2.
Thus, P = 3, Q = 2, Q^{-1} = 499122177 (modulo 998244353), and P \times Q^{-1} = 499122178 (again, modulo 998244353).

Sample Input 2

2 2

Sample Output 2

831870297

The expected number of your correct answers is 17 / 6.

Sample Input 3

3 4

Sample Output 3

770074220

The expected number of your correct answers is 169 / 35.

Sample Input 4

10 10

Sample Output 4

208827570

Sample Input 5

42 23

Sample Output 5

362936761

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;

typedef long long lint;
typedef long double louble;
template<typename T1,typename T2> inline T1 min(T1 a,T2 b){return a<b?a:b;}
template<typename T1,typename T2> inline T1 max(T1 a,T2 b){return b<a?a:b;}

namespace ae86
{
	const int bufl = 1<<15;

	char buf[bufl],*s=buf,*t=buf;

	inline int fetch()
	{
		if(s==t){t=(s=buf)+fread(buf,1,bufl,stdin);if(s==t)return EOF;}
		return *s++;
	}

	inline int ty()
	{
		int a=0,b=1,c=fetch();
		while(!isdigit(c))b^=c=='-',c=fetch();
		while(isdigit(c))a=a*10+c-48,c=fetch();
		return b?a:-a;
	}
}
using ae86::ty;

const int _ = 1000007 , mo = 998244353;

inline lint powa(lint a,lint t)
{
	lint b=1;
	while(t){if(t&1)b=b*a%mo;a=a*a%mo,t>>=1;}
	return b;
}

inline lint inva(lint a)
{
	return powa(a,mo-2);
}

lint jc[_],rjc[_];

inline void init(int n=_-1)
{
	jc[0]=jc[1]=rjc[0]=1;
	for(int i=2;i<=n;i++)jc[i]=jc[i-1]*i%mo;
	rjc[n]=inva(jc[n]);
	for(int i=n-1;i>=1;i--)rjc[i]=rjc[i+1]*(1ll+i)%mo;
}

inline lint C(int n,int m)
{
	if(n<m || n<0 || m<0)return 0;
	return jc[n]*rjc[m]%mo*rjc[n-m]%mo;
}

inline lint crx(int n,int m)
{
	return C(n+m,m);
}

int n,m;

int main()
{
	n=ty(),m=ty();
	if(n<m)swap(n,m);

	init(n+m);
	
	lint ans=0;
	for(int i=1;i<=m;i++)ans=(ans+crx(i,i)*crx(n-i,m-i)%mo)%mo;
	ans=(ans*inva(crx(n,m))%mo*inva(2)%mo+n)%mo;
	printf("%lld\n",ans);

	return 0;
}