#include <bits/stdc++.h>
using namespace std;
int main() {
  long long p, q, s[45 + 3], po;
  bool ispow = 0;
  s[0] = 1;
  cin >> p >> q;
  if (q == 1) ispow = true, po = 0;
  for (int i = 1; i <= 45; i++) {
    s[i] = s[i - 1] * 2;
    if (q == s[i]) ispow = true, po = i;
  }
  if (!ispow)
    printf("0\n");
  else {
    ++p;
    long long pp = 1, dp[45 + 3][45 + 3], ans;
    for (int i = 0; i <= 45; i++) {
      for (int j = 0; j <= 45; j++) {
        if (i == 0)
          dp[i][j] = 1;
        else if (j == 0)
          dp[i][j] = 0;
        else
          dp[i][j] = dp[i - 1][j - 1] + dp[i][j - 1];
      }
    }
    ans = 0;
    if (q == 1) --ans;
    int i;
    for (i = 0; i <= 45; i++) {
      if (p - s[i] < 0) break;
      p -= s[i];
      ans += dp[po][i];
    }
    long long l = 1, r = s[i], ppp = i;
    while (r > l) {
      long long m = (l + r) / 2;
      --ppp;
      if (p <= m - l)
        r = m;
      else {
        ans += dp[po][ppp];
        p -= (m - l + 1);
        --po;
        if (po < 0) break;
        l = m + 1;
      }
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
int main() {
  long long p, q, s[45 + 3], po;
  bool ispow = 0;
  s[0] = 1;
  cin >> p >> q;
  if (q == 1) ispow = true, po = 0;
  for (int i = 1; i <= 45; i++) {
    s[i] = s[i - 1] * 2;
    if (q == s[i]) ispow = true, po = i;
  }
  if (!ispow)
    printf("0\n");
  else {
    ++p;
    long long pp = 1, dp[45 + 3][45 + 3], ans;
    for (int i = 0; i <= 45; i++) {
      for (int j = 0; j <= 45; j++) {
        if (i == 0)
          dp[i][j] = 1;
        else if (j == 0)
          dp[i][j] = 0;
        else
          dp[i][j] = dp[i - 1][j - 1] + dp[i][j - 1];
      }
    }
    ans = 0;
    if (q == 1) --ans;
    int i;
    for (i = 0; i <= 45; i++) {
      if (p - s[i] < 0) break;
      p -= s[i];
      ans += dp[po][i];
    }
    long long l = 1, r = s[i], ppp = i;
    while (r > l) {
      long long m = (l + r) / 2;
      --ppp;
      if (p <= m - l)
        r = m;
      else {
        ans += dp[po][ppp];
        p -= (m - l + 1);
        --po;
        if (po < 0) break;
        l = m + 1;
      }
    }
    cout << ans << endl;
  }
  return 0;
}

```