#include <bits/stdc++.h>
using namespace std;
long long len[55], pCnt[55][105];
int getPower(long long x) {
  int ans = 0;
  while (x > 1) {
    if (x % 2 != 0) {
      return -1;
    }
    ans++;
    x /= 2;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long answer = 0;
  long long n, t;
  cin >> n >> t;
  n++;
  t = getPower(t);
  if (t == -1) {
    cout << "0" << '\n';
    return 0;
  }
  len[0] = 1;
  for (int i = 1; i < 50; i++) {
    len[i] = len[i - 1] + len[i - 1];
  }
  pCnt[0][0] = 1;
  for (int i = 1; i < 50; i++) {
    for (int j = 0; j < 50; j++) {
      pCnt[i][j] += pCnt[i - 1][j];
    }
    for (int j = 1; j < 50; j++) {
      pCnt[i][j] += pCnt[i - 1][j - 1];
    }
  }
  int offset = 0;
  for (int i = 0; i < 50; i++) {
    if (len[i] <= n) {
      n -= len[i];
      if (t - offset >= 0) answer += pCnt[i][t - offset];
    } else {
      break;
    }
  }
  for (int i = 49; i >= 0; i--) {
    if (len[i] <= n) {
      n -= len[i];
      if (t - offset >= 0) answer += pCnt[i][t - offset];
      offset++;
    }
  }
  if (t == 0) answer--;
  cout << answer << '\n';
}

### Prompt

Create a solution in CPP for 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 len[55], pCnt[55][105];
int getPower(long long x) {
  int ans = 0;
  while (x > 1) {
    if (x % 2 != 0) {
      return -1;
    }
    ans++;
    x /= 2;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long answer = 0;
  long long n, t;
  cin >> n >> t;
  n++;
  t = getPower(t);
  if (t == -1) {
    cout << "0" << '\n';
    return 0;
  }
  len[0] = 1;
  for (int i = 1; i < 50; i++) {
    len[i] = len[i - 1] + len[i - 1];
  }
  pCnt[0][0] = 1;
  for (int i = 1; i < 50; i++) {
    for (int j = 0; j < 50; j++) {
      pCnt[i][j] += pCnt[i - 1][j];
    }
    for (int j = 1; j < 50; j++) {
      pCnt[i][j] += pCnt[i - 1][j - 1];
    }
  }
  int offset = 0;
  for (int i = 0; i < 50; i++) {
    if (len[i] <= n) {
      n -= len[i];
      if (t - offset >= 0) answer += pCnt[i][t - offset];
    } else {
      break;
    }
  }
  for (int i = 49; i >= 0; i--) {
    if (len[i] <= n) {
      n -= len[i];
      if (t - offset >= 0) answer += pCnt[i][t - offset];
      offset++;
    }
  }
  if (t == 0) answer--;
  cout << answer << '\n';
}

```