D. Count the Arrays

Your task is to calculate the number of arrays such that:

each array contains
each element is an integer from
for each array, there is exactly one pair of equal elements;
for each array

Input

The first line contains two integers

Output

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

Examples

input

Copy

3 4

output

Copy

input

Copy

3 5

output

Copy

input

Copy

42 1337

output

Copy

806066790

input

Copy

100000 200000

output

Copy

707899035

Note

The arrays in the first example are:

n个元素有一对相同的，那么n个数中共有n-1个不同的数，从m个数中选n-1，方法数： $C(m,n-1)从n-1个不同的数中选择一个数使其在要构造的数组中出现两次，最大的数是唯一的，不能选它，所以方法数为：n-2除了重复出现的数一个在最大数的左边，一个在右边外，其他n-3个数可以出现在最大数的左边/右边，方法数为：2^n-3$

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <deque>
#include <bitset>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>
//#include <xfunctional>
#define ll              long long
#define PII             pair<int, int>
#define rep(i,a,b)      for(int  i=a;i<=b;i++)
#define dec(i,a,b)      for(int  i=a;i>=b;i--)
#define pb              push_back
#define mk              make_pair
using namespace std;
int dir1[6][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 },{ 1,1 },{ -1,1 } };
int dir2[6][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 },{ 1,-1 },{ -1,-1 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979;
const int mod = 998244353;
const int N = 1000005;
//if(x<0 || x>=r || y<0 || y>=c)

inline ll read()
{
    ll x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}
inline int add(int x, int y) {
    x += y;
    return x >= mod ? x -= mod : x;
}
inline int sub(int x, int y) {
    x -= y;
    return x < 0 ? x += mod : x;
}
inline int mul(int x, int y) {
    return 1ll * x * y % mod;
}
inline int qpow(int x, ll n) {
    int r = 1;
    while (n > 0) {
        if (n & 1) r = 1ll * r * x % mod;
        n >>= 1; x = 1ll * x * x % mod;
    }
    return r;
}
inline int Inv(int x) {
    return qpow(x, mod - 2);
}

namespace Comb {
    const int maxc = 2000000 + 5;
    int f[maxc], inv[maxc], finv[maxc];
    void init()
    {
        inv[1] = 1;
        for (int i = 2; i < maxc; i++)
            inv[i] = (mod - mod / i) * 1ll * inv[mod % i] % mod;
        f[0] = finv[0] = 1;
        for (int i = 1; i < maxc; i++)
        {
            f[i] = f[i - 1] * 1ll * i % mod;
            finv[i] = finv[i - 1] * 1ll * inv[i] % mod;
        }
    }
    int C(int n, int m)
    {
        if (m < 0 || m > n) return 0;
        return f[n] * 1ll * finv[n - m] % mod * finv[m] % mod;
    }
    int S(int n, int m)
    {
        // x_1 + x_2 + ... + x_n = m, x_i >= 0
        if (n == 0 && m == 0) return 1;
        return C(m + n - 1, n - 1);
    }
}
using Comb::C;

int main()
{
    Comb::init();
    int n, m;
    cin >> n >> m;
    ll p2=1;
    for (int i = 1; i <= n - 3; i++)
    {
        p2 *= 2;
        p2 %= mod;
    }
    ll res = ((C(m, n - 1) % mod)*(p2%mod)%mod)*((n - 2)%mod) % mod;
    cout << res << endl;
    return 0;
}