#include <bits/stdc++.h>
using namespace std;
long long n, t, res = 0;
stack<long long> s;
long long c[50][50];
void preprocess() {
  memset(c, sizeof(c), 0);
  c[0][0] = c[1][0] = c[1][1] = 1;
  for (long long i = 2; i < 50; ++i) {
    c[i][0] = 1;
    for (long long j = 1; j < i; ++j) c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
    c[i][i] = 1;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  preprocess();
  cin >> n >> t;
  ++n;
  long long p = 1, d = 1;
  while (p < t) p *= 2, ++d;
  if (p != t) {
    cout << "0";
    return 0;
  }
  p = n;
  while (p > 0) {
    s.push(p & 1);
    p /= 2;
  }
  long long m = s.size();
  while (!s.empty()) {
    long long dg = s.top();
    s.pop();
    if (d == 0) break;
    if (dg) res += c[s.size()][d--];
  }
  if (d == 0) ++res;
  if (t == 1) --res;
  cout << res;
}

### Prompt

Please create a solution in Cpp 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, res = 0;
stack<long long> s;
long long c[50][50];
void preprocess() {
  memset(c, sizeof(c), 0);
  c[0][0] = c[1][0] = c[1][1] = 1;
  for (long long i = 2; i < 50; ++i) {
    c[i][0] = 1;
    for (long long j = 1; j < i; ++j) c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
    c[i][i] = 1;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  preprocess();
  cin >> n >> t;
  ++n;
  long long p = 1, d = 1;
  while (p < t) p *= 2, ++d;
  if (p != t) {
    cout << "0";
    return 0;
  }
  p = n;
  while (p > 0) {
    s.push(p & 1);
    p /= 2;
  }
  long long m = s.size();
  while (!s.empty()) {
    long long dg = s.top();
    s.pop();
    if (d == 0) break;
    if (dg) res += c[s.size()][d--];
  }
  if (d == 0) ++res;
  if (t == 1) --res;
  cout << res;
}

```