#include <bits/stdc++.h>
using namespace std;
const int N = 50;
long long dp[N][N], n, t, ans, ol = 1;
int highbit(long long x) {
  int j;
  for (j = 0; (ol << j) <= x; ++j)
    ;
  return j - 1;
}
int main() {
  cin >> n >> t;
  dp[0][0] = 1;
  for (int i = 1; i < N; ++i) {
    dp[i][0] = 1;
    for (int j = 1; j < N; ++j) dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1];
  }
  if (t - (t & -t) != 0) {
    printf("0");
    return 0;
  }
  t = highbit(t);
  for (int i = 1; n >= (ol << i); ++i) {
    n -= ol << i;
    ans += dp[i][t];
  }
  for (int i = 0; n && t >= 0; ++i) {
    ans += dp[highbit(n)][t--];
    n -= ol << highbit(n);
  }
  cout << ans;
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
const int N = 50;
long long dp[N][N], n, t, ans, ol = 1;
int highbit(long long x) {
  int j;
  for (j = 0; (ol << j) <= x; ++j)
    ;
  return j - 1;
}
int main() {
  cin >> n >> t;
  dp[0][0] = 1;
  for (int i = 1; i < N; ++i) {
    dp[i][0] = 1;
    for (int j = 1; j < N; ++j) dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1];
  }
  if (t - (t & -t) != 0) {
    printf("0");
    return 0;
  }
  t = highbit(t);
  for (int i = 1; n >= (ol << i); ++i) {
    n -= ol << i;
    ans += dp[i][t];
  }
  for (int i = 0; n && t >= 0; ++i) {
    ans += dp[highbit(n)][t--];
    n -= ol << highbit(n);
  }
  cout << ans;
  return 0;
}

```