# Count the Arrays

• each array contains $n$ elements;
• each element is an integer from $1$ to $m$;
• for each array, there is exactly one pair of equal elements;
• for each array $a$, there exists an index $i$ such that the array is strictly ascending before the $i$-th element and strictly descending after it (formally, it means that $a_j < a_{j + 1}$, if $j < i$, and $a_j > a_{j + 1}$, if $j \ge i$).

Input

The first line contains two integers $n$ and $m$ ($2 \le n \le m \le 2 \cdot 10^5$).

Output

Print one integer — the number of arrays that meet all of the aforementioned conditions, taken modulo $998244353$.

Examples

input

3 4


output

6


input

3 5


output

10


input

42 1337


output

806066790


input

100000 200000


output

707899035


Note

The arrays in the first example are:

• $[1, 2, 1]$;
• $[1, 3, 1]$;
• $[1, 4, 1]$;
• $[2, 3, 2]$;
• $[2, 4, 2]$;
• $[3, 4, 3]$.

Solution:

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
typedef double db;
mt19937 mrand(random_device{}());
const ll mod=998244353;
int rnd(int x) { return mrand() % x;}
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}

int n,m;
ll ans=1,r=1;
int main() {
scanf("%d%d",&n,&m);
if (n==2) {
puts("0");
return 0;
}
for (int i=1;i<=m;i++) ans=ans*i%mod;
for (int i=1;i<=n-1;i++) r=r*i%mod;
for (int i=1;i<=m-n+1;i++) r=r*i%mod;
ans=ans*powmod(r,mod-2)%mod*(n-2)%mod;
ans=ans*powmod(2,n-3)%mod;
printf("%lld\n",ans);
}