#include <bits/stdc++.h>
using namespace std;
template <class T>
inline bool Up(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool Down(T &a, const T &b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
inline int getus() {
  int tmp, c;
  while (tmp = fgetc(stdin), tmp < '0' || tmp > '9')
    ;
  tmp -= '0';
  while (c = fgetc(stdin), '0' <= c && c <= '9') tmp = tmp * 10 + c - '0';
  return tmp;
}
inline int getint() {
  int tmp, c, flag;
  while (flag = fgetc(stdin), flag != '-' && (flag < '0' || flag > '9'))
    ;
  if (flag == '-')
    tmp = 0;
  else
    tmp = flag - '0';
  while (c = fgetc(stdin), '0' <= c && c <= '9') tmp = tmp * 10 + c - '0';
  return flag == '-' ? -tmp : tmp;
}
long long N, T, C[51][51];
int t;
int main() {
  cin >> N >> T;
  for (int t = 0; t <= 62; ++t)
    if ((1LL << t) > T) {
      puts("0");
      return 0;
    } else if ((1LL << t) == T) {
      T = t;
      break;
    }
  ++N;
  ++T;
  memset(C, 0, sizeof(C));
  for (int i = (0); i <= (50); ++i)
    for (int j = (0); j <= (i); ++j)
      C[i][j] = (j == 0 || j == i) ? 1 : C[i - 1][j - 1] + C[i - 1][j];
  int tot = 0;
  long long ans = 0;
  for (int i = (62); i >= (1); --i) {
    tot += (((N) >> (i)) & 1);
    if ((((N) >> (i - 1)) & 1) && T - tot >= 0) {
      ans += C[i - 1][T - tot];
    }
  }
  tot += N & 1;
  if (tot == T) ++ans;
  if (T == 1) --ans;
  cout << ans << endl;
  return 0;
}

### Prompt

Please formulate a Cpp solution 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;
template <class T>
inline bool Up(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool Down(T &a, const T &b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
inline int getus() {
  int tmp, c;
  while (tmp = fgetc(stdin), tmp < '0' || tmp > '9')
    ;
  tmp -= '0';
  while (c = fgetc(stdin), '0' <= c && c <= '9') tmp = tmp * 10 + c - '0';
  return tmp;
}
inline int getint() {
  int tmp, c, flag;
  while (flag = fgetc(stdin), flag != '-' && (flag < '0' || flag > '9'))
    ;
  if (flag == '-')
    tmp = 0;
  else
    tmp = flag - '0';
  while (c = fgetc(stdin), '0' <= c && c <= '9') tmp = tmp * 10 + c - '0';
  return flag == '-' ? -tmp : tmp;
}
long long N, T, C[51][51];
int t;
int main() {
  cin >> N >> T;
  for (int t = 0; t <= 62; ++t)
    if ((1LL << t) > T) {
      puts("0");
      return 0;
    } else if ((1LL << t) == T) {
      T = t;
      break;
    }
  ++N;
  ++T;
  memset(C, 0, sizeof(C));
  for (int i = (0); i <= (50); ++i)
    for (int j = (0); j <= (i); ++j)
      C[i][j] = (j == 0 || j == i) ? 1 : C[i - 1][j - 1] + C[i - 1][j];
  int tot = 0;
  long long ans = 0;
  for (int i = (62); i >= (1); --i) {
    tot += (((N) >> (i)) & 1);
    if ((((N) >> (i - 1)) & 1) && T - tot >= 0) {
      ans += C[i - 1][T - tot];
    }
  }
  tot += N & 1;
  if (tot == T) ++ans;
  if (T == 1) --ans;
  cout << ans << endl;
  return 0;
}

```