#include <bits/stdc++.h>
using namespace std;
long long c[60][60];
void init() {
  int i, j;
  for (i = 0; i < 50; i++) c[i][0] = 1;
  for (i = 1; i < 50; i++)
    for (j = 1; j <= i; j++) c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
}
long long get(long long n, int k) {
  int a[60], i = 0;
  long long nn = n, kk = 0;
  while (nn) {
    a[kk++] = nn % 2;
    nn >>= 1;
  }
  long long ans = 0, need = k;
  for (i = kk - 1; i >= 0; i--) {
    if (a[i]) {
      if (need >= 0) ans += c[i][need];
      need--;
    }
  }
  if (need == 0) ans++;
  if (k == 1) ans--;
  return ans;
}
int main() {
  init();
  long long n, t, i, k = 0;
  cin >> n >> t;
  long long x = 1;
  for (i = 0; i < 60; i++, x <<= 1)
    if (x == t) {
      k = i + 1;
      break;
    }
  if (!k) {
    cout << 0 << endl;
    return 0;
  }
  cout << get(n + 1, k) << endl;
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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[60][60];
void init() {
  int i, j;
  for (i = 0; i < 50; i++) c[i][0] = 1;
  for (i = 1; i < 50; i++)
    for (j = 1; j <= i; j++) c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
}
long long get(long long n, int k) {
  int a[60], i = 0;
  long long nn = n, kk = 0;
  while (nn) {
    a[kk++] = nn % 2;
    nn >>= 1;
  }
  long long ans = 0, need = k;
  for (i = kk - 1; i >= 0; i--) {
    if (a[i]) {
      if (need >= 0) ans += c[i][need];
      need--;
    }
  }
  if (need == 0) ans++;
  if (k == 1) ans--;
  return ans;
}
int main() {
  init();
  long long n, t, i, k = 0;
  cin >> n >> t;
  long long x = 1;
  for (i = 0; i < 60; i++, x <<= 1)
    if (x == t) {
      k = i + 1;
      break;
    }
  if (!k) {
    cout << 0 << endl;
    return 0;
  }
  cout << get(n + 1, k) << endl;
  return 0;
}

```