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

using ll=long long;
#define int ll

#define rng(i,a,b) for(int i=int(a);i<int(b);i++)
#define rep(i,b) rng(i,0,b)
#define gnr(i,a,b) for(int i=int(b)-1;i>=int(a);i--)
#define per(i,b) gnr(i,0,b)
#define pb push_back
#define eb emplace_back
#define a first
#define b second
#define bg begin()
#define ed end()
#define all(x) x.bg,x.ed
#ifdef LOCAL
#define dmp(x) cerr<<__LINE__<<" "<<#x<<" "<<x<<endl
#else
#define dmp(x) void(0)
#endif

template<class t,class u> void chmax(t&a,u b){if(a<b)a=b;}
template<class t,class u> void chmin(t&a,u b){if(b<a)a=b;}

template<class t> using vc=vector<t>;
template<class t> using vvc=vc<vc<t>>;

using pi=pair<int,int>;
using vi=vc<int>;

template<class t,class u>
ostream& operator<<(ostream& os,const pair<t,u>& p){
	return os<<"{"<<p.a<<","<<p.b<<"}";
}

template<class t> ostream& operator<<(ostream& os,const vc<t>& v){
	os<<"{";
	for(auto e:v)os<<e<<",";
	return os<<"}";
}

using uint=unsigned;
using ull=unsigned long long;

//initfact();
const uint mod=998244353;
//const uint mod=1000000007;
//uint mod=1;
struct mint{
	uint v;
	mint(ll vv=0){s(vv%mod+mod);}
	mint& s(uint vv){
		v=vv<mod?vv:vv-mod;
		return *this;
	}
	mint operator-()const{return mint()-*this;}
	mint& operator+=(const mint&rhs){return s(v+rhs.v);}
	mint&operator-=(const mint&rhs){return s(v+mod-rhs.v);}
	mint&operator*=(const mint&rhs){
		v=ull(v)*rhs.v%mod;
		return *this;
	}
	mint&operator/=(const mint&rhs){return *this*=rhs.inv();}
	mint operator+(const mint&rhs)const{return mint(*this)+=rhs;}
	mint operator-(const mint&rhs)const{return mint(*this)-=rhs;}
	mint operator*(const mint&rhs)const{return mint(*this)*=rhs;}
	mint operator/(const mint&rhs)const{return mint(*this)/=rhs;}
	mint pow(int n)const{
		mint res(1),x(*this);
		while(n){
			if(n&1)res*=x;
			x*=x;
			n>>=1;
		}
		return res;
	}
	mint inv()const{return pow(mod-2);}
	/*mint inv()const{
		int x,y;
		int g=extgcd(v,mod,x,y);
		assert(g==1);
		if(x<0)x+=mod;
		return mint(x);
	}*/
	friend ostream& operator<<(ostream&os,const mint&m){
		return os<<m.v;
	}
	bool operator<(const mint&r)const{return v<r.v;}
	bool operator==(const mint&r)const{return v==r.v;}
};

const int vmax=10000010;
mint fact[vmax],finv[vmax],invs[vmax];
void initfact(){
	fact[0]=1;
	rng(i,1,vmax){
		fact[i]=fact[i-1]*i;
	}
	finv[vmax-1]=fact[vmax-1].inv();
	for(int i=vmax-2;i>=0;i--){
		finv[i]=finv[i+1]*(i+1);
	}
	for(int i=vmax-1;i>=1;i--){
		invs[i]=finv[i]*fact[i-1];
	}
}
mint choose(int n,int k){
	return fact[n]*finv[n-k]*finv[k];
}
mint binom(int a,int b){
	return fact[a+b]*finv[a]*finv[b];
}
mint catalan(int n){
	return binom(n,n)-(n-1>=0?binom(n-1,n+1):0);
}

mint sub(int x,int y){
	assert(y<=x);
	mint ans=binom(x,y);
	if(y>0)ans-=binom(x+1,y-1);
	return ans;
}

signed main(){
	cin.tie(0);
	ios::sync_with_stdio(0);
	cout<<fixed<<setprecision(20);
	
	initfact();
	
	int n,a,b;cin>>n>>a>>b;
	
	mint ans=0;
	if(b==0){
		ans=1;
	}else{
		rep(i,min({a,n-b,b-1})+1){
			mint w=sub(b-1,i);
			if(i+b==n){
				if(a==i)
					ans+=w;
			}else{
				int x=n-b-i-1;
				int y=b-i-(b-a)+1;
				ans+=w*binom(x,y-1);
			}
		}
	}
	cout<<ans<<endl;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

We have two indistinguishable pieces placed on a number line. Both pieces are initially at coordinate 0. (They can occupy the same position.)

We can do the following two kinds of operations:

* Choose a piece and move it to the right (the positive direction) by 1.
* Move the piece with the smaller coordinate to the position of the piece with the greater coordinate. If two pieces already occupy the same position, nothing happens, but it still counts as doing one operation.



We want to do a total of N operations of these kinds in some order so that one of the pieces will be at coordinate A and the other at coordinate B. Find the number of ways to move the pieces to achieve it. The answer can be enormous, so compute the count modulo 998244353.

Two ways to move the pieces are considered different if and only if there exists an integer i (1 \leq i \leq N) such that the set of the coordinates occupied by the pieces after the i-th operation is different in those two ways.

Constraints

* 1 \leq N \leq 10^7
* 0 \leq A \leq B \leq N
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N A B


Output

Print the number of ways to move the pieces to achieve our objective, modulo 998244353.

Examples

Input

5 1 3


Output

4


Input

10 0 0


Output

1


Input

10 4 6


Output

197


Input

1000000 100000 200000


Output

758840509

### Response

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

using ll=long long;
#define int ll

#define rng(i,a,b) for(int i=int(a);i<int(b);i++)
#define rep(i,b) rng(i,0,b)
#define gnr(i,a,b) for(int i=int(b)-1;i>=int(a);i--)
#define per(i,b) gnr(i,0,b)
#define pb push_back
#define eb emplace_back
#define a first
#define b second
#define bg begin()
#define ed end()
#define all(x) x.bg,x.ed
#ifdef LOCAL
#define dmp(x) cerr<<__LINE__<<" "<<#x<<" "<<x<<endl
#else
#define dmp(x) void(0)
#endif

template<class t,class u> void chmax(t&a,u b){if(a<b)a=b;}
template<class t,class u> void chmin(t&a,u b){if(b<a)a=b;}

template<class t> using vc=vector<t>;
template<class t> using vvc=vc<vc<t>>;

using pi=pair<int,int>;
using vi=vc<int>;

template<class t,class u>
ostream& operator<<(ostream& os,const pair<t,u>& p){
	return os<<"{"<<p.a<<","<<p.b<<"}";
}

template<class t> ostream& operator<<(ostream& os,const vc<t>& v){
	os<<"{";
	for(auto e:v)os<<e<<",";
	return os<<"}";
}

using uint=unsigned;
using ull=unsigned long long;

//initfact();
const uint mod=998244353;
//const uint mod=1000000007;
//uint mod=1;
struct mint{
	uint v;
	mint(ll vv=0){s(vv%mod+mod);}
	mint& s(uint vv){
		v=vv<mod?vv:vv-mod;
		return *this;
	}
	mint operator-()const{return mint()-*this;}
	mint& operator+=(const mint&rhs){return s(v+rhs.v);}
	mint&operator-=(const mint&rhs){return s(v+mod-rhs.v);}
	mint&operator*=(const mint&rhs){
		v=ull(v)*rhs.v%mod;
		return *this;
	}
	mint&operator/=(const mint&rhs){return *this*=rhs.inv();}
	mint operator+(const mint&rhs)const{return mint(*this)+=rhs;}
	mint operator-(const mint&rhs)const{return mint(*this)-=rhs;}
	mint operator*(const mint&rhs)const{return mint(*this)*=rhs;}
	mint operator/(const mint&rhs)const{return mint(*this)/=rhs;}
	mint pow(int n)const{
		mint res(1),x(*this);
		while(n){
			if(n&1)res*=x;
			x*=x;
			n>>=1;
		}
		return res;
	}
	mint inv()const{return pow(mod-2);}
	/*mint inv()const{
		int x,y;
		int g=extgcd(v,mod,x,y);
		assert(g==1);
		if(x<0)x+=mod;
		return mint(x);
	}*/
	friend ostream& operator<<(ostream&os,const mint&m){
		return os<<m.v;
	}
	bool operator<(const mint&r)const{return v<r.v;}
	bool operator==(const mint&r)const{return v==r.v;}
};

const int vmax=10000010;
mint fact[vmax],finv[vmax],invs[vmax];
void initfact(){
	fact[0]=1;
	rng(i,1,vmax){
		fact[i]=fact[i-1]*i;
	}
	finv[vmax-1]=fact[vmax-1].inv();
	for(int i=vmax-2;i>=0;i--){
		finv[i]=finv[i+1]*(i+1);
	}
	for(int i=vmax-1;i>=1;i--){
		invs[i]=finv[i]*fact[i-1];
	}
}
mint choose(int n,int k){
	return fact[n]*finv[n-k]*finv[k];
}
mint binom(int a,int b){
	return fact[a+b]*finv[a]*finv[b];
}
mint catalan(int n){
	return binom(n,n)-(n-1>=0?binom(n-1,n+1):0);
}

mint sub(int x,int y){
	assert(y<=x);
	mint ans=binom(x,y);
	if(y>0)ans-=binom(x+1,y-1);
	return ans;
}

signed main(){
	cin.tie(0);
	ios::sync_with_stdio(0);
	cout<<fixed<<setprecision(20);
	
	initfact();
	
	int n,a,b;cin>>n>>a>>b;
	
	mint ans=0;
	if(b==0){
		ans=1;
	}else{
		rep(i,min({a,n-b,b-1})+1){
			mint w=sub(b-1,i);
			if(i+b==n){
				if(a==i)
					ans+=w;
			}else{
				int x=n-b-i-1;
				int y=b-i-(b-a)+1;
				ans+=w*binom(x,y-1);
			}
		}
	}
	cout<<ans<<endl;
}

```