#include<bits/stdc++.h>
//#pragma GCC optimize("O2")
//#pragma GCC optimize ("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define mp make_pair
#define pb push_back
#define pll pair<LL,LL>
#define pii pair<int,int>
#define y second
#define x first
#define LL long long 
#define ULL unsigned long long 
#define sqr(x) ((x)*(x))
#define pi acos(-1)
#define pdd pair<double,double>
#define MEM(x) memset(x,0,sizeof(x))
#define MEMS(x) memset(x,-1,sizeof(x))
using namespace std;
int dp[2][3005][3005];
int mod=998244353;
int DP(int op,int x,int y){
    if(y>x)return 0;
    if(op==0&&y==x)return 1;
    if(x==0&&y==0)return 1;
    if(y<=0)return 0;
    if(x<0)return 0;
    if(dp[op][x][y]!=-1)return dp[op][x][y];
    if(op==0){
        if(y==0)return 0;
        dp[0][x][y]=DP(1,x,y);
    }
    if(op==1){
        dp[1][x][y]=(DP(0,x,y*2)+DP(1,x-1,y-1))%mod;
    }
  //  printf("%d %d %d %d\n",op,x,y,dp[op][x][y]);
    return dp[op][x][y];
}
void solve(){   
    int n,k;
    scanf("%d %d",&n,&k);
    MEMS(dp);
    printf("%lld\n",DP(0,n,k));
}
int main(){
    int t=1;
    while(t--){
        solve();
    }
}
/*
11 4 12
1 1 1 1
AABB
.AB.
CCDD
.CD.
EEFF
.EF.
GGHH
.GH.
IIJJ
.IJ.
KKLL

12
0 0
0 -13
3 -13
3 -10
10 -10
10 10
-1 10
-1 13
-4 13
-4 10
-10 10
-10 0

*/
#include <cstdio>
#include <string>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <vector>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <tuple>
 
#define max(l,r) ((l)>(r)?(l):(r))
#define min(l,r) ((l)<(r)?(l):(r))
#define rep(i,s,l) for(int i=(int)s;i<(int)l;++i)
 
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
 
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
 
const long long INF = 1LL<<60;
const long long MOD = 1000000007;

int main() {
  int n,k;
  cin >> n >> k;
  ll ans = 0LL;
  rep(i,1,n+1) {
    rep(j,1,k+1) {
      ans += i*100+j;
    }
  }
  cout << ans << endl;
}