#include <bits/stdc++.h>
using namespace std;
const int kN = 64 + 1;
long long c[kN][kN];
void Gen() {
  c[0][0] = 1;
  for (int i = 1; i < kN; ++i) {
    c[i][0] = c[i][i] = 1;
    for (int j = 1; j < i; ++j) c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
  }
}
int cnt[1025];
void Print(int n) {
  if (n == 0) {
    cout << 0;
    return;
  }
  Print(n / 2);
  cout << n % 2;
}
vector<int> Binary(long long n) {
  vector<int> res;
  while (n) {
    res.push_back(n % 2);
    n >>= 1;
  }
  reverse(res.begin(), res.end());
  return res;
}
int main() {
  Gen();
  long long n, t;
  cin >> n >> t;
  if (t & (t - 1)) {
    cout << 0;
    return 0;
  }
  int cnt = 0;
  while (t > 1) {
    ++cnt;
    t >>= 1;
  }
  if (n == 1 && cnt == 0) {
    cout << 1;
    return 0;
  }
  int k = cnt;
  vector<int> dig = Binary(n);
  long long res = 0;
  int len = dig.size();
  for (int ind = 0; ind < dig.size() && k >= 0; ++ind) {
    if (!dig[ind]) continue;
    if (k == 0) {
      res += len - ind;
      if (ind == 0) --res;
      break;
    }
    for (int i = 0; i < len - ind - 1; ++i) res += c[i][k];
    --k;
  }
  bool ok = false;
  int sum = 0;
  for (int i = dig.size() - 1; i >= 0; --i) {
    if (dig[i] == 0) ok = true;
    sum += (ok ? dig[i] : 0);
  }
  if (sum == cnt) ++res;
  cout << res;
  return 0;
}

### 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;
const int kN = 64 + 1;
long long c[kN][kN];
void Gen() {
  c[0][0] = 1;
  for (int i = 1; i < kN; ++i) {
    c[i][0] = c[i][i] = 1;
    for (int j = 1; j < i; ++j) c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
  }
}
int cnt[1025];
void Print(int n) {
  if (n == 0) {
    cout << 0;
    return;
  }
  Print(n / 2);
  cout << n % 2;
}
vector<int> Binary(long long n) {
  vector<int> res;
  while (n) {
    res.push_back(n % 2);
    n >>= 1;
  }
  reverse(res.begin(), res.end());
  return res;
}
int main() {
  Gen();
  long long n, t;
  cin >> n >> t;
  if (t & (t - 1)) {
    cout << 0;
    return 0;
  }
  int cnt = 0;
  while (t > 1) {
    ++cnt;
    t >>= 1;
  }
  if (n == 1 && cnt == 0) {
    cout << 1;
    return 0;
  }
  int k = cnt;
  vector<int> dig = Binary(n);
  long long res = 0;
  int len = dig.size();
  for (int ind = 0; ind < dig.size() && k >= 0; ++ind) {
    if (!dig[ind]) continue;
    if (k == 0) {
      res += len - ind;
      if (ind == 0) --res;
      break;
    }
    for (int i = 0; i < len - ind - 1; ++i) res += c[i][k];
    --k;
  }
  bool ok = false;
  int sum = 0;
  for (int i = dig.size() - 1; i >= 0; --i) {
    if (dig[i] == 0) ok = true;
    sum += (ok ? dig[i] : 0);
  }
  if (sum == cnt) ++res;
  cout << res;
  return 0;
}

```