#include <bits/stdc++.h>
using namespace std;
long long f[51][51], a[51], n, t;
long long go(int step, long long value, int now) {
  if (!value || now < 0) return 0;
  if (value == a[step]) return (f[step][now]);
  if (a[step - 1] >= value) {
    long long will = go(step - 1, value, now);
    return (will);
  }
  return (f[step - 1][now] + go(step - 1, value - a[step - 1], now - 1));
}
int main() {
  scanf("%I64d%I64d", &n, &t);
  if (n == 1) {
    if (t == 1)
      printf("1\n");
    else
      printf("0\n");
    return 0;
  }
  memset(f, 0, sizeof(f));
  f[0][0] = 1;
  a[0] = 1;
  a[1] = 2;
  f[1][0] = 1;
  f[1][1] = 1;
  for (int i = 2; i <= 50; i++) {
    a[i] = a[i - 1] * 2;
    for (int j = 0; j <= 50; j++) {
      f[i][j] = f[i - 1][j];
      if (j) f[i][j] += f[i - 1][j - 1];
    }
  }
  int now = -1;
  for (int i = 0; i <= 50; i++)
    if (a[i] == t) {
      now = i;
      break;
    }
  if (now == -1) {
    printf("0\n");
    return 0;
  }
  long long ans = 0;
  for (int i = 1; i <= 50; i++)
    if (n >= a[i])
      ans += f[i][now], n -= a[i];
    else {
      ans += go(i, n, now);
      break;
    }
  printf("%I64d\n", ans);
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
long long f[51][51], a[51], n, t;
long long go(int step, long long value, int now) {
  if (!value || now < 0) return 0;
  if (value == a[step]) return (f[step][now]);
  if (a[step - 1] >= value) {
    long long will = go(step - 1, value, now);
    return (will);
  }
  return (f[step - 1][now] + go(step - 1, value - a[step - 1], now - 1));
}
int main() {
  scanf("%I64d%I64d", &n, &t);
  if (n == 1) {
    if (t == 1)
      printf("1\n");
    else
      printf("0\n");
    return 0;
  }
  memset(f, 0, sizeof(f));
  f[0][0] = 1;
  a[0] = 1;
  a[1] = 2;
  f[1][0] = 1;
  f[1][1] = 1;
  for (int i = 2; i <= 50; i++) {
    a[i] = a[i - 1] * 2;
    for (int j = 0; j <= 50; j++) {
      f[i][j] = f[i - 1][j];
      if (j) f[i][j] += f[i - 1][j - 1];
    }
  }
  int now = -1;
  for (int i = 0; i <= 50; i++)
    if (a[i] == t) {
      now = i;
      break;
    }
  if (now == -1) {
    printf("0\n");
    return 0;
  }
  long long ans = 0;
  for (int i = 1; i <= 50; i++)
    if (n >= a[i])
      ans += f[i][now], n -= a[i];
    else {
      ans += go(i, n, now);
      break;
    }
  printf("%I64d\n", ans);
}

```