#include <bits/stdc++.h>
using namespace std;
int f(int n) {
  int r = 0;
  while (n) (n &= (n - 1)), r++;
  return 1 << (r - 1);
}
long long cr[100][100];
long long C(int n, int k) {
  if (n == 0 && k == 0) return 1;
  if (n < 0 || k < 0) return 0;
  if (cr[n][k] == -1) {
    cr[n][k] = C(n - 1, k) + C(n - 1, k - 1);
  }
  return cr[n][k];
}
long long int count(long long int max, int bitsCount) {
  if (bitsCount == 0) return 1;
  if (max == 0) return 0;
  long long int b = 1;
  int c = 0;
  while (b <= max) (b *= 2), c++;
  if (c < bitsCount) return 0;
  b /= 2;
  return C(c - 1, bitsCount) + count(max - b, bitsCount - 1);
}
void write() {
  const int m = 100;
  int a[m + 2][m + 2];
  for (int i = 1; i <= m; i++) a[1][i] = 0;
  a[1][m + 1] = 1;
  for (int i = 2; i <= m + 1; i++) {
    for (int j = 1; j <= m + 1; j++) {
      if (j == 1)
        a[i][j] = a[i - 1][j + 1];
      else {
        if (j == m + 1)
          a[i][j] = a[i - 1][j - 1];
        else {
          a[i][j] = a[i - 1][j - 1] ^ a[i - 1][j + 1];
        }
      }
    }
    char ch[2] = {'.', '#'};
    cout << setw(3) << i << ": ";
    int ss = 0;
    for (int j = 1; j <= m + 1; j++) cout << ch[a[i][j]];
    cout << ss << " " << f(i);
    cout << endl;
  }
}
int main() {
  std::map<long long int, int> mm;
  for (int i = 0; i < 60; i++) mm[1ll << i] = i;
  memset(cr, -1, sizeof cr);
  long long int n, t;
  cin >> n >> t;
  auto iter = mm.find(t);
  if (iter == mm.end()) return cout << 0, 0;
  auto rr = count(n + 1, iter->second + 1);
  cout << (iter->second == 0 ? rr - 1 : rr);
}

### Prompt

Please create a solution in CPP 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;
int f(int n) {
  int r = 0;
  while (n) (n &= (n - 1)), r++;
  return 1 << (r - 1);
}
long long cr[100][100];
long long C(int n, int k) {
  if (n == 0 && k == 0) return 1;
  if (n < 0 || k < 0) return 0;
  if (cr[n][k] == -1) {
    cr[n][k] = C(n - 1, k) + C(n - 1, k - 1);
  }
  return cr[n][k];
}
long long int count(long long int max, int bitsCount) {
  if (bitsCount == 0) return 1;
  if (max == 0) return 0;
  long long int b = 1;
  int c = 0;
  while (b <= max) (b *= 2), c++;
  if (c < bitsCount) return 0;
  b /= 2;
  return C(c - 1, bitsCount) + count(max - b, bitsCount - 1);
}
void write() {
  const int m = 100;
  int a[m + 2][m + 2];
  for (int i = 1; i <= m; i++) a[1][i] = 0;
  a[1][m + 1] = 1;
  for (int i = 2; i <= m + 1; i++) {
    for (int j = 1; j <= m + 1; j++) {
      if (j == 1)
        a[i][j] = a[i - 1][j + 1];
      else {
        if (j == m + 1)
          a[i][j] = a[i - 1][j - 1];
        else {
          a[i][j] = a[i - 1][j - 1] ^ a[i - 1][j + 1];
        }
      }
    }
    char ch[2] = {'.', '#'};
    cout << setw(3) << i << ": ";
    int ss = 0;
    for (int j = 1; j <= m + 1; j++) cout << ch[a[i][j]];
    cout << ss << " " << f(i);
    cout << endl;
  }
}
int main() {
  std::map<long long int, int> mm;
  for (int i = 0; i < 60; i++) mm[1ll << i] = i;
  memset(cr, -1, sizeof cr);
  long long int n, t;
  cin >> n >> t;
  auto iter = mm.find(t);
  if (iter == mm.end()) return cout << 0, 0;
  auto rr = count(n + 1, iter->second + 1);
  cout << (iter->second == 0 ? rr - 1 : rr);
}

```