#include <bits/stdc++.h>
using namespace std;
struct debugger {
  template <typename T>
  debugger& operator,(const T& v) {
    cerr << v << " ";
    return *this;
  }
} dbg;
void debugarr(int* arr, int n) {
  cout << "[";
  for (int i = 0; i < n; i++) cout << arr[i] << " ";
  cout << "]" << endl;
}
long long fun1(long long mat[50][50], long long poi, long long n, long long t) {
  if (t == 1) return 1;
  if (poi < 0) return 0;
  int lg = 0;
  long long c = t;
  while (c > 0) {
    c /= 2;
    lg++;
  }
  lg--;
  long long ans = 0;
  long long p = ((long long)1) << (poi - 1);
  if (n == p * 2) return mat[poi][lg];
  if (n < p) {
    ans = fun1(mat, poi - 1, n, t);
  } else if (n == p) {
    return mat[poi - 1][lg];
  } else {
    ans = fun1(mat, poi - 1, n % p, t / 2);
    ans += mat[poi - 1][lg];
  }
  return ans;
}
long long fun(long long n, long long t) {
  long long mat[50][50] = {{0}};
  mat[0][0] = 1;
  for (int i = 1; i <= 48; i++) {
    for (int j = 1; j <= 48; j++) mat[i][j] = mat[i - 1][j - 1];
    for (int j = 0; j <= 48; j++) mat[i][j] += mat[i - 1][j];
  }
  int lg = 0;
  long long temp = t;
  while (temp > 0) {
    temp /= 2;
    lg++;
  }
  lg--;
  int poi = 1;
  long long ans = 0;
  while (1) {
    if ((((long long)1) << poi) > n) break;
    n -= ((long long)1) << poi;
    ans += mat[poi][lg];
    poi++;
  }
  if (n != 0) ans += fun1(mat, poi, n, t);
  return ans;
}
int main() {
  long long n, t;
  cin >> n >> t;
  long long c = t;
  int f = 0;
  for (int i = 0; i <= 50 - 1; i++) {
    if (c == (((long long)1) << i)) f = 1;
  }
  if (f == 0) {
    cout << 0 << endl;
    return 0;
  }
  long long ans = fun(n, t);
  cout << ans << endl;
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

Maxim loves to fill in a matrix in a special manner. Here is a pseudocode of filling in a matrix of size (m + 1) × (m + 1):

<image>

Maxim asks you to count, how many numbers m (1 ≤ m ≤ n) are there, such that the sum of values in the cells in the row number m + 1 of the resulting matrix equals t.

Expression (x xor y) means applying the operation of bitwise excluding "OR" to numbers x and y. The given operation exists in all modern programming languages. For example, in languages C++ and Java it is represented by character "^", in Pascal — by "xor".

Input

A single line contains two integers n and t (1 ≤ n, t ≤ 1012, t ≤ n + 1).

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.

Output

In a single line print a single integer — the answer to the problem. 

Examples

Input

1 1


Output

1


Input

3 2


Output

1


Input

3 3


Output

0


Input

1000000000000 1048576


Output

118606527258

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
struct debugger {
  template <typename T>
  debugger& operator,(const T& v) {
    cerr << v << " ";
    return *this;
  }
} dbg;
void debugarr(int* arr, int n) {
  cout << "[";
  for (int i = 0; i < n; i++) cout << arr[i] << " ";
  cout << "]" << endl;
}
long long fun1(long long mat[50][50], long long poi, long long n, long long t) {
  if (t == 1) return 1;
  if (poi < 0) return 0;
  int lg = 0;
  long long c = t;
  while (c > 0) {
    c /= 2;
    lg++;
  }
  lg--;
  long long ans = 0;
  long long p = ((long long)1) << (poi - 1);
  if (n == p * 2) return mat[poi][lg];
  if (n < p) {
    ans = fun1(mat, poi - 1, n, t);
  } else if (n == p) {
    return mat[poi - 1][lg];
  } else {
    ans = fun1(mat, poi - 1, n % p, t / 2);
    ans += mat[poi - 1][lg];
  }
  return ans;
}
long long fun(long long n, long long t) {
  long long mat[50][50] = {{0}};
  mat[0][0] = 1;
  for (int i = 1; i <= 48; i++) {
    for (int j = 1; j <= 48; j++) mat[i][j] = mat[i - 1][j - 1];
    for (int j = 0; j <= 48; j++) mat[i][j] += mat[i - 1][j];
  }
  int lg = 0;
  long long temp = t;
  while (temp > 0) {
    temp /= 2;
    lg++;
  }
  lg--;
  int poi = 1;
  long long ans = 0;
  while (1) {
    if ((((long long)1) << poi) > n) break;
    n -= ((long long)1) << poi;
    ans += mat[poi][lg];
    poi++;
  }
  if (n != 0) ans += fun1(mat, poi, n, t);
  return ans;
}
int main() {
  long long n, t;
  cin >> n >> t;
  long long c = t;
  int f = 0;
  for (int i = 0; i <= 50 - 1; i++) {
    if (c == (((long long)1) << i)) f = 1;
  }
  if (f == 0) {
    cout << 0 << endl;
    return 0;
  }
  long long ans = fun(n, t);
  cout << ans << endl;
  return 0;
}

```