#include <bits/stdc++.h>
using namespace std;
long long n, t, ans;
long long f[100][100];
long long one = 1;
long long work(long long n, long long t, long long p) {
  if (n == 0) return 0;
  long long u = 1, v = 0, ans = 0;
  while (u + u <= n) {
    u += u;
    v++;
  }
  long long e = t / p, g = 1, h = 0;
  while (g + g <= e) {
    g += g;
    h++;
  }
  if (e > 0) ans = f[v][h] + work(n - u, t, p + p);
  return ans;
}
int main() {
  f[0][0] = 1;
  for (int i = 0; i < 50; i++) {
    for (int j = 0; j <= i; j++) {
      f[i + 1][j] += f[i][j];
      f[i + 1][j + 1] += f[i][j];
    }
    f[i + 1][1]--;
    f[i + 1][0]++;
  }
  cin >> n >> t;
  n++;
  ans = 0;
  long long tt = t, one = 1;
  bool b = true;
  while (tt > one) {
    if (tt & one) b = false;
    tt >>= one;
  }
  if (b) {
    ans = work(n, t, 1);
    if (t == one) ans--;
  }
  cout << ans << 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 n, t, ans;
long long f[100][100];
long long one = 1;
long long work(long long n, long long t, long long p) {
  if (n == 0) return 0;
  long long u = 1, v = 0, ans = 0;
  while (u + u <= n) {
    u += u;
    v++;
  }
  long long e = t / p, g = 1, h = 0;
  while (g + g <= e) {
    g += g;
    h++;
  }
  if (e > 0) ans = f[v][h] + work(n - u, t, p + p);
  return ans;
}
int main() {
  f[0][0] = 1;
  for (int i = 0; i < 50; i++) {
    for (int j = 0; j <= i; j++) {
      f[i + 1][j] += f[i][j];
      f[i + 1][j + 1] += f[i][j];
    }
    f[i + 1][1]--;
    f[i + 1][0]++;
  }
  cin >> n >> t;
  n++;
  ans = 0;
  long long tt = t, one = 1;
  bool b = true;
  while (tt > one) {
    if (tt & one) b = false;
    tt >>= one;
  }
  if (b) {
    ans = work(n, t, 1);
    if (t == one) ans--;
  }
  cout << ans << endl;
  return 0;
}

```