#include <bits/stdc++.h>
using namespace std;
long long s[45][45];
long long nbit[45];
int main() {
  s[0][0] = 1;
  for (int i = 1; i <= 42; i++) {
    s[i][0] = s[i][i] = 1;
    for (int j = 1; j < i; j++) s[i][j] = s[i - 1][j - 1] + s[i - 1][j];
  }
  long long n, t;
  cin >> n >> t;
  n += 2;
  long long tt = t;
  bool feasible = true;
  long long bitcnt = 0;
  while (tt > 0) {
    if (tt & 1 && tt != 1) {
      feasible = false;
      break;
    }
    tt >>= 1;
    bitcnt++;
  }
  long long nn = n;
  long long nptr = 0;
  while (nn > 0) {
    nbit[nptr++] = nn & 1;
    nn >>= 1;
  }
  if (feasible) {
    long long answer = 0;
    long long onecnt = 0;
    for (int ni = nptr - 1; ni >= 0; ni--) {
      if (nbit[ni] == 1) answer += s[ni][bitcnt - onecnt];
      onecnt += nbit[ni];
    }
    if (bitcnt == 1) answer--;
    cout << answer << endl;
  } else
    cout << 0 << endl;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 s[45][45];
long long nbit[45];
int main() {
  s[0][0] = 1;
  for (int i = 1; i <= 42; i++) {
    s[i][0] = s[i][i] = 1;
    for (int j = 1; j < i; j++) s[i][j] = s[i - 1][j - 1] + s[i - 1][j];
  }
  long long n, t;
  cin >> n >> t;
  n += 2;
  long long tt = t;
  bool feasible = true;
  long long bitcnt = 0;
  while (tt > 0) {
    if (tt & 1 && tt != 1) {
      feasible = false;
      break;
    }
    tt >>= 1;
    bitcnt++;
  }
  long long nn = n;
  long long nptr = 0;
  while (nn > 0) {
    nbit[nptr++] = nn & 1;
    nn >>= 1;
  }
  if (feasible) {
    long long answer = 0;
    long long onecnt = 0;
    for (int ni = nptr - 1; ni >= 0; ni--) {
      if (nbit[ni] == 1) answer += s[ni][bitcnt - onecnt];
      onecnt += nbit[ni];
    }
    if (bitcnt == 1) answer--;
    cout << answer << endl;
  } else
    cout << 0 << endl;
  return 0;
}

```