#include <bits/stdc++.h>
using namespace std;
long long dp[41][41];
void pre() {
  for (int i = 0; i < 41; ++i) dp[0][i] = 0;
  dp[0][0] = 1;
  for (int i = 1; i < 41; ++i) {
    dp[i][0] = dp[i - 1][0];
    for (int j = 1; j < 41; ++j) dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1];
  }
}
long long n, t, cur;
int findInd() {
  cur = 0;
  int ret = 0;
  for (int i = 0; i < 41; ++i) {
    cur += (1LL << i);
    if (cur > n) {
      cur -= (1LL << i);
      ret = i - 1;
      break;
    }
  }
  return ret;
}
void calc() {
  int po = 0;
  for (int i = 0; i < 41; ++i) {
    if ((1LL << i) == t) {
      po = i;
      break;
    }
  }
  int ind = findInd();
  long long ans = 0;
  for (int i = 0; i <= ind; ++i) {
    ans += dp[i][po];
  }
  if (po == 0) ans--;
  long long c = n - cur;
  while (c && ind >= 0 && po >= 0) {
    if (c >= (1LL << (ind))) {
      c -= (1LL << (ind));
      ans += dp[ind][po];
      ind--;
      po--;
    } else {
      ind--;
    }
  }
  cout << ans << endl;
}
int main() {
  cin >> n >> t;
  ++n;
  if ((t & (t - 1)) != 0) {
    cout << "0" << endl;
    return 0;
  }
  pre();
  calc();
  return 0;
}

### Prompt

Your task is to create 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 dp[41][41];
void pre() {
  for (int i = 0; i < 41; ++i) dp[0][i] = 0;
  dp[0][0] = 1;
  for (int i = 1; i < 41; ++i) {
    dp[i][0] = dp[i - 1][0];
    for (int j = 1; j < 41; ++j) dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1];
  }
}
long long n, t, cur;
int findInd() {
  cur = 0;
  int ret = 0;
  for (int i = 0; i < 41; ++i) {
    cur += (1LL << i);
    if (cur > n) {
      cur -= (1LL << i);
      ret = i - 1;
      break;
    }
  }
  return ret;
}
void calc() {
  int po = 0;
  for (int i = 0; i < 41; ++i) {
    if ((1LL << i) == t) {
      po = i;
      break;
    }
  }
  int ind = findInd();
  long long ans = 0;
  for (int i = 0; i <= ind; ++i) {
    ans += dp[i][po];
  }
  if (po == 0) ans--;
  long long c = n - cur;
  while (c && ind >= 0 && po >= 0) {
    if (c >= (1LL << (ind))) {
      c -= (1LL << (ind));
      ans += dp[ind][po];
      ind--;
      po--;
    } else {
      ind--;
    }
  }
  cout << ans << endl;
}
int main() {
  cin >> n >> t;
  ++n;
  if ((t & (t - 1)) != 0) {
    cout << "0" << endl;
    return 0;
  }
  pre();
  calc();
  return 0;
}

```