#include <bits/stdc++.h>
using namespace std;
long long dp[55][55], n, t, p;
int tot, i, j;
long long work(long long a, long long b) {
  if (a == 1) {
    if (b == 0)
      return 1;
    else
      return 0;
  }
  if (b < 0) return 0;
  long long x = 1, tot = 0;
  while (x <= a) {
    x *= 2;
    tot++;
  }
  x /= 2;
  tot--;
  if (x == a) return dp[tot][b] + (b == 0);
  if (x * 2 - 1 == a) return dp[tot + 1][b];
  return dp[tot][b] + (b == 0) + work(a - x, b - 1);
}
int main() {
  cin >> n >> t;
  n++;
  p = 1;
  while (p < t) {
    p *= 2;
    tot++;
  }
  if (p != t) {
    cout << 0;
    return 0;
  }
  dp[1][0] = 1;
  for (i = 2; i <= 50; i++)
    for (j = 0; j <= 50; j++)
      if (j)
        dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1];
      else
        dp[i][j] = dp[i - 1][j] + 1;
  cout << work(n, tot) - (t == 1);
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
long long dp[55][55], n, t, p;
int tot, i, j;
long long work(long long a, long long b) {
  if (a == 1) {
    if (b == 0)
      return 1;
    else
      return 0;
  }
  if (b < 0) return 0;
  long long x = 1, tot = 0;
  while (x <= a) {
    x *= 2;
    tot++;
  }
  x /= 2;
  tot--;
  if (x == a) return dp[tot][b] + (b == 0);
  if (x * 2 - 1 == a) return dp[tot + 1][b];
  return dp[tot][b] + (b == 0) + work(a - x, b - 1);
}
int main() {
  cin >> n >> t;
  n++;
  p = 1;
  while (p < t) {
    p *= 2;
    tot++;
  }
  if (p != t) {
    cout << 0;
    return 0;
  }
  dp[1][0] = 1;
  for (i = 2; i <= 50; i++)
    for (j = 0; j <= 50; j++)
      if (j)
        dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1];
      else
        dp[i][j] = dp[i - 1][j] + 1;
  cout << work(n, tot) - (t == 1);
  return 0;
}

```