#include <bits/stdc++.h>
using namespace std;
const double PI = 2.0 * acos(0.0);
const double EPS = 1e-6;
long long C[42][42];
void precall() {
  C[0][0] = C[1][0] = C[1][1] = 1LL;
  for (int r = 2; r < 42; r++) C[r][0] = 1LL;
  for (int r = 2; r < 42; r++)
    for (int c = 1; c <= r; c++) C[r][c] = C[r - 1][c] + C[r - 1][c - 1];
}
int one_bit_count(long long x) {
  int res = 0;
  while (x > 0) {
    if (x & 1) res++;
    x >>= 1;
  }
  return res;
}
int main() {
  precall();
  long long n, t;
  scanf("%I64d %I64d", &n, &t);
  if (one_bit_count(t) > 1) return puts("0"), 0;
  int need = 0;
  for (int i = 0; i < 42; i++)
    if ((((t) >> (i)) & 1)) need = i + 1;
  long long res = 0LL;
  int one = one_bit_count(n + 1);
  if (one == need) res++;
  for (int i = 0; i < 42; i++)
    if ((((n + 1) >> (i)) & 1)) {
      one--;
      if (need >= one) res += C[i][need - one];
    }
  if (t == 1LL) res--;
  printf("%I64d\n", res);
  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;
const double PI = 2.0 * acos(0.0);
const double EPS = 1e-6;
long long C[42][42];
void precall() {
  C[0][0] = C[1][0] = C[1][1] = 1LL;
  for (int r = 2; r < 42; r++) C[r][0] = 1LL;
  for (int r = 2; r < 42; r++)
    for (int c = 1; c <= r; c++) C[r][c] = C[r - 1][c] + C[r - 1][c - 1];
}
int one_bit_count(long long x) {
  int res = 0;
  while (x > 0) {
    if (x & 1) res++;
    x >>= 1;
  }
  return res;
}
int main() {
  precall();
  long long n, t;
  scanf("%I64d %I64d", &n, &t);
  if (one_bit_count(t) > 1) return puts("0"), 0;
  int need = 0;
  for (int i = 0; i < 42; i++)
    if ((((t) >> (i)) & 1)) need = i + 1;
  long long res = 0LL;
  int one = one_bit_count(n + 1);
  if (one == need) res++;
  for (int i = 0; i < 42; i++)
    if ((((n + 1) >> (i)) & 1)) {
      one--;
      if (need >= one) res += C[i][need - one];
    }
  if (t == 1LL) res--;
  printf("%I64d\n", res);
  return 0;
}

```