#include <bits/stdc++.h>
using namespace std;
const int max_n = 200;
long long n, t;
vector<long long> a[max_n];
int find(long long l) {
  long long p = 1;
  int res = 0;
  while (1) {
    if (l == p) break;
    if (p > l) {
      res--;
      break;
    }
    p *= 2;
    res++;
  }
  return res;
}
long long pow(int x) {
  if (x == 0) return 1;
  long long res = pow(x / 2);
  res *= res;
  if (x % 2 == 1) res *= 2;
  return res;
}
int main() {
  cin >> n >> t;
  bool b = 1;
  long long tmp = t;
  while (tmp > 1) {
    if (tmp % 2 == 1) b = 0;
    tmp /= 2;
  }
  if (b) {
    a[0].push_back(1);
    for (int i = 1; i < max_n; i++) {
      a[i].push_back(1);
      for (int j = 0; j < a[i - 1].size() - 1; j++) {
        a[i].push_back(a[i - 1][j] + a[i - 1][j + 1]);
      }
      a[i].push_back(1);
    }
    n = n + 1;
    long long res = 0;
    int p = find(t);
    int lp = find(n);
    for (int i = 1; i < lp; i++)
      if (a[i].size() > p) {
        res += (long long)(a[i][p]);
      }
    long long ptr = (long long)(pow(lp)) - 1;
    while (ptr < n) {
      lp = find(n - ptr);
      if (p >= 0 and a[lp].size() > p) res += (long long)a[lp][p];
      ptr += (long long)(pow(lp));
      p--;
    }
    cout << res << endl;
  } else
    printf("0\n");
  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 max_n = 200;
long long n, t;
vector<long long> a[max_n];
int find(long long l) {
  long long p = 1;
  int res = 0;
  while (1) {
    if (l == p) break;
    if (p > l) {
      res--;
      break;
    }
    p *= 2;
    res++;
  }
  return res;
}
long long pow(int x) {
  if (x == 0) return 1;
  long long res = pow(x / 2);
  res *= res;
  if (x % 2 == 1) res *= 2;
  return res;
}
int main() {
  cin >> n >> t;
  bool b = 1;
  long long tmp = t;
  while (tmp > 1) {
    if (tmp % 2 == 1) b = 0;
    tmp /= 2;
  }
  if (b) {
    a[0].push_back(1);
    for (int i = 1; i < max_n; i++) {
      a[i].push_back(1);
      for (int j = 0; j < a[i - 1].size() - 1; j++) {
        a[i].push_back(a[i - 1][j] + a[i - 1][j + 1]);
      }
      a[i].push_back(1);
    }
    n = n + 1;
    long long res = 0;
    int p = find(t);
    int lp = find(n);
    for (int i = 1; i < lp; i++)
      if (a[i].size() > p) {
        res += (long long)(a[i][p]);
      }
    long long ptr = (long long)(pow(lp)) - 1;
    while (ptr < n) {
      lp = find(n - ptr);
      if (p >= 0 and a[lp].size() > p) res += (long long)a[lp][p];
      ptr += (long long)(pow(lp));
      p--;
    }
    cout << res << endl;
  } else
    printf("0\n");
  return 0;
}

```