#include <bits/stdc++.h>
#pragma warning(disable : 4996)
long long MX = 1000000000000LL;
long long C[60][60];
int p[60];
int main() {
  long long n, t, x, Res = 0;
  int c = 0, m = 0, i, j;
  C[0][0] = 1;
  for (i = 1; i < 60; i++) {
    C[i][0] = 1;
    for (j = 1; j < 60; j++) {
      C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
    }
  }
  scanf("%lld%lld", &n, &t);
  x = 1;
  while (1) {
    if (t == x) break;
    if (x > MX) break;
    x *= 2;
    c++;
  }
  if (x > MX) {
    printf("0\n");
    return 0;
  }
  c++;
  while (n) {
    p[m++] = n % 2;
    n /= 2;
  }
  p[1]++;
  for (i = 1; p[i] == 2; i++) {
    p[i + 1]++;
    p[i] = 0;
  }
  if (p[m + 1]) m += 2;
  if (p[m]) m++;
  for (i = m - 1; i >= 0; i--) {
    if (p[i] == 1) {
      Res += C[i][c];
      c--;
      if (c < 0) break;
    }
  }
  if (t == 1) Res--;
  printf("%lld\n", Res);
}

### 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>
#pragma warning(disable : 4996)
long long MX = 1000000000000LL;
long long C[60][60];
int p[60];
int main() {
  long long n, t, x, Res = 0;
  int c = 0, m = 0, i, j;
  C[0][0] = 1;
  for (i = 1; i < 60; i++) {
    C[i][0] = 1;
    for (j = 1; j < 60; j++) {
      C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
    }
  }
  scanf("%lld%lld", &n, &t);
  x = 1;
  while (1) {
    if (t == x) break;
    if (x > MX) break;
    x *= 2;
    c++;
  }
  if (x > MX) {
    printf("0\n");
    return 0;
  }
  c++;
  while (n) {
    p[m++] = n % 2;
    n /= 2;
  }
  p[1]++;
  for (i = 1; p[i] == 2; i++) {
    p[i + 1]++;
    p[i] = 0;
  }
  if (p[m + 1]) m += 2;
  if (p[m]) m++;
  for (i = m - 1; i >= 0; i--) {
    if (p[i] == 1) {
      Res += C[i][c];
      c--;
      if (c < 0) break;
    }
  }
  if (t == 1) Res--;
  printf("%lld\n", Res);
}

```