#include <iostream>
#include <string>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <tuple>
#include <map>
#include<set>
#include<queue>
#include<stack>
#include <unordered_set>
#include<thread>
#include<bits/stdc++.h>
#include <numbers>

#include <atcoder/all>
#include <cstdio>

#pragma GCC target("avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")

//if(a < 0 || h <= a || b < 0 || w <= b)return;


using namespace std;
using namespace atcoder;
using ll = long long;
using ld = long double;
using ull = unsigned long long;
using mint = modint998244353;
using mint1 = modint1000000007;
//using VL = vector<ll>;
template<typename T> using pq = priority_queue<T>;//降順?(最大取り出し)
template<typename T> using pqg = priority_queue<T, vector<T>, greater<T>>;//昇順?(最小取り出し)
template<typename T> using vector2 = vector<vector<T>>;
template<typename T> using vector3 = vector<vector<vector<T>>>;
template<typename T> using vector4 = vector<vector<vector<vector<T>>>>;
template<typename T> using vector5 = vector<vector<vector<vector<vector<T>>>>>;
template<typename T> using vector6 = vector<vector<vector<vector<vector<vector<T>>>>>>;
template<typename T> using pairs = pair<T,T>;
#define rep(i, n) for (ll i = 0; i < ll(n); i++)
#define rep1(i,n) for(int i = 1;i <= int(n);i++)
#define repm(i, m, n) for (int i = (m); (i) < int(n);(i)++)
#define repmr(i, m, n) for (int i = (m) - 1; (i) >= int(n);(i)--)
#define rep0(i,n) for(int i = n - 1;i >= 0;i--)
#define rep01(i,n) for(int i = n;i >= 1;i--)



// NのK乗根N < 2^64, K <= 64
uint64_t kth_root(uint64_t N, uint64_t K) {
    assert(K >= 1);
    if (N <= 1 || K == 1) return N;
    if (K >= 64) return 1;
    if (N == uint64_t(-1)) --N;
    
    auto mul = [&](uint64_t x, uint64_t y) -> uint64_t {
        if (x < UINT_MAX && y < UINT_MAX) return x * y;
        if (x == uint64_t(-1) || y == uint64_t(-1)) return uint64_t(-1);
        return (x <= uint64_t(-1) / y ? x * y : uint64_t(-1));
    };
    auto power = [&](uint64_t x, uint64_t k) -> uint64_t {
        if (k == 0) return 1ULL;
        uint64_t res = 1ULL;
        while (k) {
            if (k & 1) res = mul(res, x);
            x = mul(x, x);
            k >>= 1;
        }
        return res;
    };
    
    uint64_t res;
    if (K == 2) res = sqrtl(N) - 1;
    else if (K == 3) res = cbrt(N) - 1;
    else res = pow(N, nextafter(1 / double(K), 0));
    while (power(res + 1, K) <= N) ++res;
    return res;
}
// ユークリッドの互除法による最大公約数算出
ll GCD(ll a,ll b){
    if(b == 0)return a;
    return GCD(b, a % b);
}
//拡張ユークリッドの互除法による(ax + by = GCD(a,b))を満たすx,yの算出
pair<long long, long long> extgcd(long long a, long long b) {
  if (b == 0) return make_pair(1, 0);
  long long x, y;
  tie(y, x) = extgcd(b, a % b);
  y -= a / b * x;
  return make_pair(x, y);
}
struct UnionFind {
    vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2
    
    UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化
        for(int i = 0; i < N; i++) par[i] = i;
    }
    
    int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }
    
    void unite(int x, int y) { // xとyの木を併合
        int rx = root(x); //xの根をrx
        int ry = root(y); //yの根をry
        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
        par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
    }
    
    bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }
};
ll n;

//座標圧縮
vector<ll> Ccomp(vector<ll> a){
    vector<ll> b = a;
    sort(b.begin(),b.end());
    b.erase(unique(b.begin(),b.end()),b.end());//ダブり消去
    vector<ll> rtn;
    rep(j,a.size()){
        ll pb = lower_bound(b.begin(),b.end(),a[j]) - b.begin();
        rtn.push_back(pb);
    }
    return rtn;
}
/// ここから////////////////////////////////////////////




using F = pairs<ll>;
using S = pairs<ll>;
string s;

ll modPow(ll a, ll n, ll mod) { if(mod==1) return 0;ll ret = 1; ll p = a % mod; while (n) { if (n & 1) ret = ret * p % mod; p = p * p % mod; n >>= 1; } return ret; }
void cincout(){
  ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
  cout<< fixed << setprecision(15);
}
//seg,遅延segの設定-----ここから
S op(S a,S b){
    return a;
    
}//何を求めるか(最大値とか)

S e(){return {1,n};}

S mapping (F a,S b){
    if(a.second == 0)return b;
    else return a;
}//遅延処理

F composition (F a,F b){
    if(a.second == 0)return b;
    return a;
}//遅延中の枝にさらに処理

F id(){return {0,0};}//遅延のモノイド

vector<int> Op(vector<int> a,vector<int> b){a.insert(a.end(),b.begin(),b.end()); return a;}

vector<int> E(){return vector<int> (0);}

//segここまで
string abc = "abcdefghijklmnopqrstuvwxyz";
string Labc = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

ll mod = ll(1e9) + 7;
ll INF  =ll(2e18);

vector<ll> movey = {-1,0,1,0},movex = {0,1,0,-1};
bool outc(ll nowy,ll nowx,ll h,ll w){
    if(nowy < 0 || nowy >= h || nowx < 0 || nowx >= w){
        return 0;
    }
    return 1;
}

int main() {
    cincout();
    ll n,m;
    cin >> n >> m;
    vector<mint> fact(2*m+10,1);
    for(ll j = 2;j <= 2*m;j++){
        fact[j] = fact[j-1] * j;
    }
    if(2*m-1<n){
        cout << 0 << endl;
        return 0;
    }
    mint ans = 0;
    for(int j = 1;j <= m;j++){
        if(2 * j - 1 < n)continue;
        ans += fact[2*(j-1)] / fact[n-1] / fact[2*(j-1)-(n-1)];
    }
    cout << ans.val() << endl;
    return 0;
}