#include <bits/stdc++.h>
using namespace std;
const int N(55);
long long d[N];
long long c[N][N];
long long n, t;
int getPow(long long t) {
  int x = 0;
  while (t != 1) {
    if (t % 2) return -1;
    t /= 2;
    ++x;
  }
  return x;
}
long long cal(int x, long long n, int p) {
  if (x == -1 || p == -1) return 0;
  if (d[x] <= n) return c[x][p] + cal(x - 1, n - d[x], p - 1);
  return cal(x - 1, n, p);
}
int main() {
  cin >> n >> t;
  d[0] = 1;
  for (int i = int(1); i < int(N); ++i) d[i] = d[i - 1] * 2;
  for (int i = int(0); i < int(N); ++i) {
    c[i][0] = 1;
    for (int j = int(1); j < int(i + 1); ++j)
      c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
  }
  int p = getPow(t);
  long long ans = 0;
  if (p != -1) {
    ++n;
    t = 1;
    for (int i = int(0); i < int(N); ++i) {
      if (t <= n) {
        ans += c[i][p];
        n -= t;
      }
      t *= 2;
    }
    ans += cal(N - 1, n, p);
    if (p == 0) --ans;
  }
  cout << ans << 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;
const int N(55);
long long d[N];
long long c[N][N];
long long n, t;
int getPow(long long t) {
  int x = 0;
  while (t != 1) {
    if (t % 2) return -1;
    t /= 2;
    ++x;
  }
  return x;
}
long long cal(int x, long long n, int p) {
  if (x == -1 || p == -1) return 0;
  if (d[x] <= n) return c[x][p] + cal(x - 1, n - d[x], p - 1);
  return cal(x - 1, n, p);
}
int main() {
  cin >> n >> t;
  d[0] = 1;
  for (int i = int(1); i < int(N); ++i) d[i] = d[i - 1] * 2;
  for (int i = int(0); i < int(N); ++i) {
    c[i][0] = 1;
    for (int j = int(1); j < int(i + 1); ++j)
      c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
  }
  int p = getPow(t);
  long long ans = 0;
  if (p != -1) {
    ++n;
    t = 1;
    for (int i = int(0); i < int(N); ++i) {
      if (t <= n) {
        ans += c[i][p];
        n -= t;
      }
      t *= 2;
    }
    ans += cal(N - 1, n, p);
    if (p == 0) --ans;
  }
  cout << ans << endl;
  return 0;
}

```