#include <bits/stdc++.h>
using namespace std;
inline void read(int &first) {
  first = 0;
  while (1) {
    char ch = getchar();
    if (ch == ' ' || ch == '\n') break;
    first = first * 10 + ch - '0';
  }
}
inline void write(int first) {
  char wr[12];
  int k = 0;
  if (!first) ++k, wr[k] = '0';
  while (first) {
    ++k;
    wr[k] = char(first % 10 + '0');
    first /= 10;
  }
  for (int i = k; i >= 1; --i) putchar(wr[i]);
}
long long n, t, ans, z, C[41][41];
long long solve(long long n) {
  vector<int> a;
  long long ans = 0;
  int kol = 0;
  while (n) {
    a.push_back(n & 1ll);
    n >>= 1ll;
  }
  reverse((a).begin(), (a).end());
  for (int i = 0; i < a.size(); ++i) {
    if (!a[i]) continue;
    if (kol > z) break;
    ans += C[a.size() - i - 1][z - kol];
    kol += a[i];
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> t;
  for (int i = 0; i <= 40; ++i) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; ++j) C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
  }
  if (__builtin_popcountll(t) != 1) return cout << 0, 0;
  z = __builtin_popcountll(t - 1) + 1;
  cout << solve(n + 2) - (t == 1);
}

### 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;
inline void read(int &first) {
  first = 0;
  while (1) {
    char ch = getchar();
    if (ch == ' ' || ch == '\n') break;
    first = first * 10 + ch - '0';
  }
}
inline void write(int first) {
  char wr[12];
  int k = 0;
  if (!first) ++k, wr[k] = '0';
  while (first) {
    ++k;
    wr[k] = char(first % 10 + '0');
    first /= 10;
  }
  for (int i = k; i >= 1; --i) putchar(wr[i]);
}
long long n, t, ans, z, C[41][41];
long long solve(long long n) {
  vector<int> a;
  long long ans = 0;
  int kol = 0;
  while (n) {
    a.push_back(n & 1ll);
    n >>= 1ll;
  }
  reverse((a).begin(), (a).end());
  for (int i = 0; i < a.size(); ++i) {
    if (!a[i]) continue;
    if (kol > z) break;
    ans += C[a.size() - i - 1][z - kol];
    kol += a[i];
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> t;
  for (int i = 0; i <= 40; ++i) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; ++j) C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
  }
  if (__builtin_popcountll(t) != 1) return cout << 0, 0;
  z = __builtin_popcountll(t - 1) + 1;
  cout << solve(n + 2) - (t == 1);
}

```