#include <bits/stdc++.h>
using namespace std;
vector<int> v;
long long res;
int W;
long long C[111][111];
long long c(int n, int m) {
  if (m < 0 || m > n) return 0;
  return C[n][m];
}
void go(int pos, int num, bool sm = 0) {
  if (sm || pos == v.size()) {
    int xx = v.size() - pos;
    res += c(xx, W - num);
  } else {
    for (int i = 0; i <= v[pos]; i++) go(pos + 1, num + i, i < v[pos]);
  }
}
int main() {
  for (int i = 0; i < 100; i++) {
    C[i][i] = C[i][0] = 1;
    for (int j = 1; j < i; j++) C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
  }
  long long n, t;
  cin >> n >> t;
  long long x = t & (t - 1);
  if (x)
    puts("0");
  else {
    int num = 0;
    while (t) num++, t >>= 1;
    W = num;
    res = 0;
    n++;
    while (n) {
      v.push_back(n % 2);
      n /= 2;
    }
    reverse(v.begin(), v.end());
    go(0, 0);
    if (num == 1) res--;
    cout << res << endl;
  }
  return 0;
}

### 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;
vector<int> v;
long long res;
int W;
long long C[111][111];
long long c(int n, int m) {
  if (m < 0 || m > n) return 0;
  return C[n][m];
}
void go(int pos, int num, bool sm = 0) {
  if (sm || pos == v.size()) {
    int xx = v.size() - pos;
    res += c(xx, W - num);
  } else {
    for (int i = 0; i <= v[pos]; i++) go(pos + 1, num + i, i < v[pos]);
  }
}
int main() {
  for (int i = 0; i < 100; i++) {
    C[i][i] = C[i][0] = 1;
    for (int j = 1; j < i; j++) C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
  }
  long long n, t;
  cin >> n >> t;
  long long x = t & (t - 1);
  if (x)
    puts("0");
  else {
    int num = 0;
    while (t) num++, t >>= 1;
    W = num;
    res = 0;
    n++;
    while (n) {
      v.push_back(n % 2);
      n /= 2;
    }
    reverse(v.begin(), v.end());
    go(0, 0);
    if (num == 1) res--;
    cout << res << endl;
  }
  return 0;
}

```