using namespace std;

// 高精度整数：十进制 1e9 进制，小端存储
struct BigInt {
    vector<int> d;
    BigInt() : d(1, 0) {}
    BigInt(int x) {
        if (x == 0) d.push_back(0);
        else {
            while (x) { d.push_back(x % 1000000000); x /= 1000000000; }
        }
    }
    void trim() { while (d.size() > 1 && d.back() == 0) d.pop_back(); }

    BigInt operator+(const BigInt& o) const {
        BigInt r;
        r.d.assign(max(d.size(), o.d.size()) + 1, 0);
        long long carry = 0;
        for (size_t i = 0; i < r.d.size(); i++) {
            long long s = carry;
            if (i < d.size()) s += d[i];
            if (i < o.d.size()) s += o.d[i];
            r.d[i] = (int)(s % 1000000000);
            carry = s / 1000000000;
        }
        r.trim();
        return r;
    }
    // 前提：*this >= o
    BigInt operator-(const BigInt& o) const {
        BigInt r;
        r.d.assign(d.size(), 0);
        long long borrow = 0;
        for (size_t i = 0; i < d.size(); i++) {
            long long s = (long long)d[i] - borrow - (i < o.d.size() ? o.d[i] : 0);
            if (s < 0) { s += 1000000000; borrow = 1; } else borrow = 0;
            r.d[i] = (int)s;
        }
        r.trim();
        return r;
    }
    void print() const {
        printf("%d", d.back());
        for (int i = (int)d.size() - 2; i >= 0; i--) printf("%09d", d[i]);
        printf("\n");
    }
};

int main() {
    int k, w;
    scanf("%d%d", &k, &w);
    int B = 1 << k;          // 2^k（k<=9，B<=512）
    int n = w / k, m = w % k;

    // 递推杨辉三角，只需保留上一行；快照第 B-1 行与第 B-2^m 行
    vector<BigInt> prev(1, BigInt(1)), rowB1, rowBmin;
    for (int top = 0; top <= B - 1; top++) {
        vector<BigInt> cur(top + 1, BigInt(0));
        cur[0] = BigInt(1);
        for (int j = 1; j <= top; j++)
            cur[j] = prev[j - 1] + (j < (int)prev.size() ? prev[j] : BigInt(0));
        if (top == B - 1) rowB1 = cur;
        if (m > 0 && top == B - (1 << m)) rowBmin = cur;
        prev = move(cur);
    }

    BigInt ans;
    for (int d = 2; d <= min(n, B - 1); d++) ans = ans + rowB1[d];

    if (m > 0) {
        // 补上 d = n+1 位的贡献：C(B-1, n+1) - C(B-2^m, n+1)
        BigInt a = (n + 1 <= B - 1) ? rowB1[n + 1] : BigInt(0);
        BigInt b = (n + 1 <= B - (1 << m)) ? rowBmin[n + 1] : BigInt(0);
        ans = ans + (a - b);
    }
    ans.print();
    return 0;
}