#include <bits/stdc++.h>
using namespace std;
const int MAXN = 45;
long long dp[MAXN][MAXN];
int main() {
  memset(dp, 0, sizeof(dp));
  dp[0][0] = 1;
  int i, j, p;
  for (i = 1; i < MAXN; i++) {
    dp[i][0] += dp[i - 1][0];
    for (j = 1; j <= i; j++) {
      dp[i][j] += dp[i - 1][j] + dp[i - 1][j - 1];
    }
  }
  long long n, t;
  while (cin >> n >> t) {
    long long lans = 0, one = 1;
    for (i = 0; i < MAXN; i++)
      if (t == (one << i)) break;
    if (i == MAXN) {
      puts("0");
      continue;
    }
    t = i;
    p = 1;
    while (n >= (one << p)) {
      lans += dp[p][t];
      n -= (one << p++);
    }
    for (i = MAXN - 1; i >= 0 && t >= 0; i--) {
      if (n & (one << i)) lans += dp[i][t--];
    }
    cout << lans << endl;
  }
  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 MAXN = 45;
long long dp[MAXN][MAXN];
int main() {
  memset(dp, 0, sizeof(dp));
  dp[0][0] = 1;
  int i, j, p;
  for (i = 1; i < MAXN; i++) {
    dp[i][0] += dp[i - 1][0];
    for (j = 1; j <= i; j++) {
      dp[i][j] += dp[i - 1][j] + dp[i - 1][j - 1];
    }
  }
  long long n, t;
  while (cin >> n >> t) {
    long long lans = 0, one = 1;
    for (i = 0; i < MAXN; i++)
      if (t == (one << i)) break;
    if (i == MAXN) {
      puts("0");
      continue;
    }
    t = i;
    p = 1;
    while (n >= (one << p)) {
      lans += dp[p][t];
      n -= (one << p++);
    }
    for (i = MAXN - 1; i >= 0 && t >= 0; i--) {
      if (n & (one << i)) lans += dp[i][t--];
    }
    cout << lans << endl;
  }
  return 0;
}

```