#include <bits/stdc++.h>
using namespace std;
static const int INF = 500000000;
template <class T>
void debug(T a, T b) {
  for (; a != b; ++a) cerr << *a << ' ';
  cerr << endl;
}
long long int dp[100][100];
long long int gett(int h, int p, long long int n) {
  if (n == 0 || p < 0) return 0;
  if ((1ll << h) <= n) return dp[h][p];
  long long int half = (1ll << (h - 1));
  if (n <= half)
    return gett(h - 1, p, n);
  else
    return gett(h - 1, p, half) + gett(h - 1, p - 1, n - half);
}
long long int solve(long long int n, int p) {
  long long int sum = 1;
  int h = 1;
  long long int res = 0;
  while (sum < n) {
    res += gett(h, p, n - sum);
    sum += (1ll << h);
    ++h;
  }
  return res;
}
int main() {
  dp[0][0] = 1;
  for (int i = 0; i < 50; ++i)
    for (int j = 0; j < 50; ++j)
      dp[i + 1][j] = dp[i][j] + (j >= 1 ? dp[i][j - 1] : 0);
  long long int n, t;
  cin >> n >> t;
  ++n;
  int p = 0;
  while ((1ll << p) < t) ++p;
  if ((1ll << p) != t) {
    puts("0");
    return 0;
  }
  cout << solve(n, p) << endl;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
static const int INF = 500000000;
template <class T>
void debug(T a, T b) {
  for (; a != b; ++a) cerr << *a << ' ';
  cerr << endl;
}
long long int dp[100][100];
long long int gett(int h, int p, long long int n) {
  if (n == 0 || p < 0) return 0;
  if ((1ll << h) <= n) return dp[h][p];
  long long int half = (1ll << (h - 1));
  if (n <= half)
    return gett(h - 1, p, n);
  else
    return gett(h - 1, p, half) + gett(h - 1, p - 1, n - half);
}
long long int solve(long long int n, int p) {
  long long int sum = 1;
  int h = 1;
  long long int res = 0;
  while (sum < n) {
    res += gett(h, p, n - sum);
    sum += (1ll << h);
    ++h;
  }
  return res;
}
int main() {
  dp[0][0] = 1;
  for (int i = 0; i < 50; ++i)
    for (int j = 0; j < 50; ++j)
      dp[i + 1][j] = dp[i][j] + (j >= 1 ? dp[i][j - 1] : 0);
  long long int n, t;
  cin >> n >> t;
  ++n;
  int p = 0;
  while ((1ll << p) < t) ++p;
  if ((1ll << p) != t) {
    puts("0");
    return 0;
  }
  cout << solve(n, p) << endl;
  return 0;
}

```