#include <bits/stdc++.h>
using namespace std;
long long n, t, ll1, tmp;
int i, j, g;
long long dp[100][100], b[100];
int main() {
  cin >> n >> t;
  b[1] = 1;
  for (i = 2; i <= 50; i++) b[i] = b[i - 1] << 1;
  for (i = 1; i <= 50; i++)
    if (t == b[i]) break;
  if (t != b[i]) {
    cout << 0 << endl;
    return 0;
  }
  t = i;
  n += 2;
  ll1 = 1;
  for (i = 50; i >= 0; i--) {
    tmp = ll1 << i;
    if (n & tmp) {
      dp[i][g] = 1;
      g++;
    }
  }
  for (i = 50; i >= 0; i--)
    for (j = 0; j <= 60; j++)
      dp[i][j] = dp[i][j] + dp[i + 1][j - 1] + dp[i + 1][j];
  if (t == 1) dp[0][t]--;
  cout << dp[0][t] << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
long long n, t, ll1, tmp;
int i, j, g;
long long dp[100][100], b[100];
int main() {
  cin >> n >> t;
  b[1] = 1;
  for (i = 2; i <= 50; i++) b[i] = b[i - 1] << 1;
  for (i = 1; i <= 50; i++)
    if (t == b[i]) break;
  if (t != b[i]) {
    cout << 0 << endl;
    return 0;
  }
  t = i;
  n += 2;
  ll1 = 1;
  for (i = 50; i >= 0; i--) {
    tmp = ll1 << i;
    if (n & tmp) {
      dp[i][g] = 1;
      g++;
    }
  }
  for (i = 50; i >= 0; i--)
    for (j = 0; j <= 60; j++)
      dp[i][j] = dp[i][j] + dp[i + 1][j - 1] + dp[i + 1][j];
  if (t == 1) dp[0][t]--;
  cout << dp[0][t] << endl;
  return 0;
}

```