#include <bits/stdc++.h>
using namespace std;
long long C[55][55];
long long g(int t, int x, long long n) {
  if (t < 0 || n == 0) return 0;
  if (x == 0) {
    if (t == 0) return 1;
    return 0;
  }
  if (n <= (1ll << (x - 1)))
    return g(t, x - 1, n);
  else {
    return C[x - 1][t] + g(t - 1, x - 1, n - (1ll << (x - 1)));
  }
}
long long f(int t, long long n) {
  if (t < 0) return 0;
  long long res = 0;
  long long two;
  int i;
  for (i = 1, two = 2; two <= n; i++, two <<= 1) {
    res += C[i][t];
    n -= two;
  }
  return res + g(t, i, n);
}
int bit(long long x) {
  int res = 0;
  x >>= 1;
  while (x) {
    x >>= 1;
    res++;
  }
  return res;
}
int main() {
  int i, j, k;
  long long n, t;
  cin >> n >> t;
  if (t & (t - 1)) {
    puts("0");
    return 0;
  }
  for (i = 0; i < 55; i++) {
    C[i][0] = 1;
    for (j = 1; j <= i; j++) C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
  }
  cout << f(bit(t), n);
  return 0;
}

### Prompt

Generate 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;
long long C[55][55];
long long g(int t, int x, long long n) {
  if (t < 0 || n == 0) return 0;
  if (x == 0) {
    if (t == 0) return 1;
    return 0;
  }
  if (n <= (1ll << (x - 1)))
    return g(t, x - 1, n);
  else {
    return C[x - 1][t] + g(t - 1, x - 1, n - (1ll << (x - 1)));
  }
}
long long f(int t, long long n) {
  if (t < 0) return 0;
  long long res = 0;
  long long two;
  int i;
  for (i = 1, two = 2; two <= n; i++, two <<= 1) {
    res += C[i][t];
    n -= two;
  }
  return res + g(t, i, n);
}
int bit(long long x) {
  int res = 0;
  x >>= 1;
  while (x) {
    x >>= 1;
    res++;
  }
  return res;
}
int main() {
  int i, j, k;
  long long n, t;
  cin >> n >> t;
  if (t & (t - 1)) {
    puts("0");
    return 0;
  }
  for (i = 0; i < 55; i++) {
    C[i][0] = 1;
    for (j = 1; j <= i; j++) C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
  }
  cout << f(bit(t), n);
  return 0;
}

```