#include <bits/stdc++.h>
using namespace std;
long long a[200005], b[200005];
map<pair<long long, long long>, long long> mp;
long long fun(long long n, long long k) {
  pair<long long, long long> p = {n, k};
  if (mp.find(p) != mp.end()) return mp[p];
  if (n == 0) {
    return mp[p] = 0;
  }
  if (n == 1) {
    if (k >= 1) {
      return mp[p] = 0;
    } else {
      return mp[p] = 1;
    }
  }
  long long ans = 0;
  ans += fun((n + 1) / 2, k);
  if (k > 0) {
    ans += fun(n / 2, k - 1);
  }
  return mp[p] = ans;
}
int32_t main() {
  a[0] = 1;
  for (long long i = 1; i <= 40; i++) a[i] = a[i - 1] * i;
  long long n, t;
  cin >> n >> t;
  if ((t & (t - 1))) {
    cout << 0;
  } else {
    long long x = log2(t) + 1;
    cout << fun(n + 2, x) - fun(2, x);
  }
}

### Prompt

Generate 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;
long long a[200005], b[200005];
map<pair<long long, long long>, long long> mp;
long long fun(long long n, long long k) {
  pair<long long, long long> p = {n, k};
  if (mp.find(p) != mp.end()) return mp[p];
  if (n == 0) {
    return mp[p] = 0;
  }
  if (n == 1) {
    if (k >= 1) {
      return mp[p] = 0;
    } else {
      return mp[p] = 1;
    }
  }
  long long ans = 0;
  ans += fun((n + 1) / 2, k);
  if (k > 0) {
    ans += fun(n / 2, k - 1);
  }
  return mp[p] = ans;
}
int32_t main() {
  a[0] = 1;
  for (long long i = 1; i <= 40; i++) a[i] = a[i - 1] * i;
  long long n, t;
  cin >> n >> t;
  if ((t & (t - 1))) {
    cout << 0;
  } else {
    long long x = log2(t) + 1;
    cout << fun(n + 2, x) - fun(2, x);
  }
}

```