#include <bits/stdc++.h>
using namespace std;
long long mas[50][50];
long long step[50];
long long res[50];
int main() {
  step[0] = 1;
  for (int i = 1; i < 50; i++) step[i] = step[i - 1] * 2;
  long long n, t;
  cin >> n >> t;
  int st = -1;
  for (int i = 0; i < 50; i++) {
    if (t == step[i]) {
      st = i;
      break;
    }
  }
  if (st == -1) {
    cout << 0 << endl;
    return 0;
  }
  mas[0][0] = 1;
  for (int i = 1; i < 50; i++) {
    for (int j = 0; j < 50; j++) {
      mas[i][j] = mas[i - 1][j];
      if (j > 0) mas[i][j] += mas[i - 1][j - 1];
    }
  }
  long long r = 0, oldr = 0;
  for (int i = 1;; i++) {
    r += step[i];
    if (r > n) break;
    for (int j = 0; j < 50; j++) res[j] += mas[i][j];
    oldr = r;
  }
  n -= oldr;
  int cnt = 0;
  for (int i = 49; i >= 0; i--) {
    if (n >= step[i]) {
      for (int j = 0; j + cnt < 50; j++) res[j + cnt] += mas[i][j];
      cnt++;
      n -= step[i];
    }
  }
  cout << res[st] << endl;
  return 0;
}

### Prompt

Your task is to create 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 mas[50][50];
long long step[50];
long long res[50];
int main() {
  step[0] = 1;
  for (int i = 1; i < 50; i++) step[i] = step[i - 1] * 2;
  long long n, t;
  cin >> n >> t;
  int st = -1;
  for (int i = 0; i < 50; i++) {
    if (t == step[i]) {
      st = i;
      break;
    }
  }
  if (st == -1) {
    cout << 0 << endl;
    return 0;
  }
  mas[0][0] = 1;
  for (int i = 1; i < 50; i++) {
    for (int j = 0; j < 50; j++) {
      mas[i][j] = mas[i - 1][j];
      if (j > 0) mas[i][j] += mas[i - 1][j - 1];
    }
  }
  long long r = 0, oldr = 0;
  for (int i = 1;; i++) {
    r += step[i];
    if (r > n) break;
    for (int j = 0; j < 50; j++) res[j] += mas[i][j];
    oldr = r;
  }
  n -= oldr;
  int cnt = 0;
  for (int i = 49; i >= 0; i--) {
    if (n >= step[i]) {
      for (int j = 0; j + cnt < 50; j++) res[j + cnt] += mas[i][j];
      cnt++;
      n -= step[i];
    }
  }
  cout << res[st] << endl;
  return 0;
}

```