/*

     题意：转换就是求n位数字，总和为s/2的方案数

     DP+高精度：状态转移方程：dp[cur^1][k+j] = dp[cur^1][k+j] + dp[cur][k];

                 高精度直接拿JayYe的:)

     异或运算的规则:

     0⊕0=0,0⊕1=1

     1⊕0=1,1⊕1=0

     口诀：相同取0，相异取1

 */

 #include <cstdio>

 #include <cstring>

 #include <string>

 #include <iostream>

 #include <algorithm>

 using namespace std;

 const int numlen = ;

 const int numbit = ;

 const int addbit = ;

 const int maxn = numlen/numbit + ;

 struct bign {

     int len, s[numlen];

     bign() {

         memset(s, , sizeof(s));

         len = ;

     }

     bign(int num) { *this = num; }

     bign(const char *num) { *this = num; }

     bign operator = (const int num) {

         char s[numlen];

         sprintf(s, "%d", num);

         *this = s;

         return *this;

     }

     bign operator = (const char *num){

         int clen = strlen(num);

         while(clen >  && num[] == '')    num++, clen--;

         len = ;

         for(int i = clen-;i >= ; i -= numbit) {

             int top = min(numbit, i+), mul = ;

             s[len] = ;

             for(int j = ;j < top; j++) {

                 s[len] += (num[i-j]-'')*mul;

                 mul *= ;

             }

             len++;

         }

         deal();

         return *this;

     }

     void deal() {

         while(len >  && !s[len-]) len--;

     }

     bign operator + (const bign &a) const {

         bign ret;

         ret.len = ;

         int top = max(len, a.len), add = ;

         for(int i = ;add || i < top; i++) {

             int now = add;

             if(i < len) now += s[i];

             if(i < a.len)   now += a.s[i];

             ret.s[ret.len++] = now%addbit;

             add = now/addbit;

         }

         return ret;

     }

     bign operator * (const bign &a)const {

         bign ret;

         ret.len = len + a.len;

         for(int i = ;i < len; i++) {

             int pre = ;

             for(int j = ;j < a.len; j++) {

                 int now = s[i]*a.s[j] + pre;

                 pre = ;

                 ret.s[i+j] += now;

                 if(ret.s[i+j] >= addbit) {

                     pre = ret.s[i+j]/addbit;

                     ret.s[i+j] -= pre*addbit;

                 }

             }

             if(pre) ret.s[i+a.len] = pre;

         }

         ret.deal();

         return ret;

     }

 }dp[][];

 istream& operator >> (istream &in, bign &x) {

     string s;

     in >> s;

     x = s.c_str();

     return in;

 }

 ostream& operator << (ostream &out, const bign &x) {

     printf("%d", x.s[x.len-]);

     for(int i = x.len-;i >= ; i--)    printf("%04d", x.s[i]);

     return out;

 }

 int main(void)        //URAL 1036 Lucky Tickets

 {

     //freopen ("W.in", "r", stdin);

     int n, s;

     while (scanf ("%d%d", &n, &s) == )

     {

         if (s & )    {puts ("");    continue;}

         dp[][] = ;

         int cur = ;

         for (int i=; i<=n; ++i)

         {

             for (int j=; j<=s/; ++j)    dp[cur^][j] = ;

             for (int j=; j<=; ++j)

             {

                 for (int k=; k+j<=s/; ++k)

                     dp[cur^][k+j] = dp[cur^][k+j] + dp[cur][k];

             }

             cur ^= ;

         }

         cout << dp[cur][s/] * dp[cur][s/] << endl;

     }

     return ;

 }