#include <bits/stdc++.h>
using namespace std;
int popcount(long long n) {
  int ans = 0;
  while (n) ans += n % 2, n /= 2;
  return ans;
}
vector<int> bin;
void cut(long long n) {
  bin.clear();
  while (n) {
    bin.push_back(n % 2);
    n /= 2;
  }
  reverse(bin.begin(), bin.end());
}
long long n, t;
long long memo[100][100][2];
long long solve(int idx, int bits, bool less) {
  if (idx == (int)bin.size()) return bits == t;
  long long &ret = memo[idx][bits][less];
  if (ret != -1) return ret;
  ret = 0;
  for (int i = 0; i < 2; i++) {
    if (less) {
      ret += solve(idx + 1, bits + (i == 1), less);
    } else {
      if (i < bin[idx]) {
        ret += solve(idx + 1, bits + (i == 1), 1);
      } else if (i == bin[idx]) {
        ret += solve(idx + 1, bits + (i == 1), 0);
      }
    }
  }
  return ret;
}
int main() {
  cin >> n >> t;
  if (t & (t - 1)) {
    cout << 0 << endl;
    return 0;
  }
  int cnt = 0;
  while (t) {
    t /= 2;
    cnt++;
  }
  t = cnt;
  cut(n + 1);
  memset(memo, -1, sizeof memo);
  long long ans = solve(0, 0, 0);
  cut(1);
  memset(memo, -1, sizeof memo);
  ans -= solve(0, 0, 0);
  cout << ans << endl;
}

### 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;
int popcount(long long n) {
  int ans = 0;
  while (n) ans += n % 2, n /= 2;
  return ans;
}
vector<int> bin;
void cut(long long n) {
  bin.clear();
  while (n) {
    bin.push_back(n % 2);
    n /= 2;
  }
  reverse(bin.begin(), bin.end());
}
long long n, t;
long long memo[100][100][2];
long long solve(int idx, int bits, bool less) {
  if (idx == (int)bin.size()) return bits == t;
  long long &ret = memo[idx][bits][less];
  if (ret != -1) return ret;
  ret = 0;
  for (int i = 0; i < 2; i++) {
    if (less) {
      ret += solve(idx + 1, bits + (i == 1), less);
    } else {
      if (i < bin[idx]) {
        ret += solve(idx + 1, bits + (i == 1), 1);
      } else if (i == bin[idx]) {
        ret += solve(idx + 1, bits + (i == 1), 0);
      }
    }
  }
  return ret;
}
int main() {
  cin >> n >> t;
  if (t & (t - 1)) {
    cout << 0 << endl;
    return 0;
  }
  int cnt = 0;
  while (t) {
    t /= 2;
    cnt++;
  }
  t = cnt;
  cut(n + 1);
  memset(memo, -1, sizeof memo);
  long long ans = solve(0, 0, 0);
  cut(1);
  memset(memo, -1, sizeof memo);
  ans -= solve(0, 0, 0);
  cout << ans << endl;
}

```