#include <bits/stdc++.h>
using namespace std;
long long int c[77][77];
long long int n, t;
void do1() {
  long long int i, j;
  c[0][0] = 1;
  for (i = 1; i <= 50; i++)
    for (j = 0; j <= i; j++) c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
}
long long int do2() {
  long long int i, j, k, l, r, m, p, q;
  j = 0;
  p = 0;
  n++;
  n++;
  for (i = 1; i < t; i <<= 1) j++;
  j++;
  if (i > t) return 0;
  for (k = 49; k >= 0; k--)
    if (n & (1ll << k)) {
      p += c[k][j];
      j--;
      if (j < 0) break;
    }
  if (t == 1) p--;
  return p;
}
int main() {
  cin >> n >> t;
  do1();
  cout << do2();
  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;
long long int c[77][77];
long long int n, t;
void do1() {
  long long int i, j;
  c[0][0] = 1;
  for (i = 1; i <= 50; i++)
    for (j = 0; j <= i; j++) c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
}
long long int do2() {
  long long int i, j, k, l, r, m, p, q;
  j = 0;
  p = 0;
  n++;
  n++;
  for (i = 1; i < t; i <<= 1) j++;
  j++;
  if (i > t) return 0;
  for (k = 49; k >= 0; k--)
    if (n & (1ll << k)) {
      p += c[k][j];
      j--;
      if (j < 0) break;
    }
  if (t == 1) p--;
  return p;
}
int main() {
  cin >> n >> t;
  do1();
  cout << do2();
  return 0;
}

```