Problem: In Republic of AtCoder, Snuke Chameleons (Family: Chamaeleonidae, Genus: Bartaberia) are very popular pets. Ringo keeps N Snuke Chameleons in a cage.

A Snuke Chameleon that has not eaten anything is blue. It changes its color according to the following rules:

* A Snuke Chameleon that is blue will change its color to red when the number of red balls it has eaten becomes strictly larger than the number of blue balls it has eaten.
* A Snuke Chameleon that is red will change its color to blue when the number of blue balls it has eaten becomes strictly larger than the number of red balls it has eaten.



Initially, every Snuke Chameleon had not eaten anything. Ringo fed them by repeating the following process K times:

* Grab either a red ball or a blue ball.
* Throw that ball into the cage. Then, one of the chameleons eats it.



After Ringo threw in K balls, all the chameleons were red. We are interested in the possible ways Ringo could have thrown in K balls. How many such ways are there? Find the count modulo 998244353. Here, two ways to throw in balls are considered different when there exists i such that the color of the ball that are thrown in the i-th throw is different.

Constraints

* 1 \leq N,K \leq 5 \times 10^5
* N and K are integers.

Input

Input is given from Standard Input in the following format:


N K


Output

Print the possible ways Ringo could have thrown in K balls, modulo 998244353.

Examples

Input

2 4


Output

7


Input

3 7


Output

57


Input

8 3


Output

0


Input

8 10


Output

46


Input

123456 234567


Output

857617983

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define meow(args...) fprintf(stderr, args)
template<class T1, class T2> inline bool cmin(T1 &a, const T2 &b) {return b<a?(a=b, true):false;}
template<class T1, class T2> inline bool cmax(T1 &a, const T2 &b) {return a<b?(a=b, true):false;}
template<class Type> Type read() {
	Type a;
	bool b;
	unsigned char c;
	while(c=getchar()-48, (c>9)&(c!=253));
	for(a=(b=c==253)?0:c; (c=getchar()-48)<=9; a=a*10+c);
	return b?-a:a;
}
int (*rd)()=read<int>;
typedef unsigned u32;
typedef long long s64;
typedef unsigned long long u64;
const u32 P=998244353;
u32 &inc(u32 &a, u32 b) {return (a+=b)<P?a:(a-=P);}
u32 &dec(u32 &a, u32 b) {return (a-=b)&0x80000000?(a+=P):a;}
u32 sum(u32 a, u32 b) {return inc(a, b);}
u32 dif(u32 a, u32 b) {return dec(a, b);}
u64 power(u64 a, int b) {
	u64 ans=1;
	for(; b; a=a*a%P, b/=2) if(b&1) ans=ans*a%P;
	return ans;
}

const int N=5e5+5;
u32 fac[N], ifac[N];
u64 com(int n, int r) {
	return n<0||r<0||r>n?0:(u64)fac[n]*ifac[r]%P*ifac[n-r]%P;
}
u64 sol(int x, int y, int l) {
	return com(x+y, x)-com(x+y, x+l)+P;
}
int main() {
	int n=rd(), k=rd();
	u32 ans=0;
	fac[0]=1;
	for(int i=1; i<=k; ++i) fac[i]=(u64)fac[i-1]*i%P;
	ifac[k]=power(fac[k], P-2);
	for(int i=k; i--; ) ifac[i]=ifac[i+1]*(i+1llu)%P;
	for(int r=std::max(n, (k+1)/2); r<=k; ++r)
		ans=(ans+sol(r, k-r-(r==k-r), r-n+1))%P;
	printf("%u\n", ans);
	return 0;
}