#include <bits/stdc++.h>
using namespace std;
int popcount(long long x) {
  int r = 0;
  while (x) {
    if (x & 1) r++;
    x >>= 1;
  }
  return r;
}
long long n, t;
long long dp[66][66][2];
long long rec(int i, int x, bool lt) {
  if (x < 0) return 0;
  if (dp[i + 1][x][lt] != -1) return dp[i + 1][x][lt];
  long long r = 0;
  if (i == -1)
    r = x == 0;
  else
    for (int(b) = 0; (b) < (int)(2); ++(b))
      if (lt || !((n >> i & 1) == 0 && b == 1)) {
        r += rec(i - 1, x - b, lt || ((n >> i & 1) && b == 0));
      }
  return dp[i + 1][x][lt] = r;
}
int main() {
  cin >> n >> t;
  if (popcount(t) != 1) {
    cout << "0" << endl;
    return 0;
  }
  int log2 = 0;
  while ((1LL << log2) != t) log2++;
  memset(dp, -1, sizeof(dp));
  ++n;
  long long r = rec(62, log2 + 1, false);
  if (log2 == 0) r--;
  cout << r << endl;
  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;
int popcount(long long x) {
  int r = 0;
  while (x) {
    if (x & 1) r++;
    x >>= 1;
  }
  return r;
}
long long n, t;
long long dp[66][66][2];
long long rec(int i, int x, bool lt) {
  if (x < 0) return 0;
  if (dp[i + 1][x][lt] != -1) return dp[i + 1][x][lt];
  long long r = 0;
  if (i == -1)
    r = x == 0;
  else
    for (int(b) = 0; (b) < (int)(2); ++(b))
      if (lt || !((n >> i & 1) == 0 && b == 1)) {
        r += rec(i - 1, x - b, lt || ((n >> i & 1) && b == 0));
      }
  return dp[i + 1][x][lt] = r;
}
int main() {
  cin >> n >> t;
  if (popcount(t) != 1) {
    cout << "0" << endl;
    return 0;
  }
  int log2 = 0;
  while ((1LL << log2) != t) log2++;
  memset(dp, -1, sizeof(dp));
  ++n;
  long long r = rec(62, log2 + 1, false);
  if (log2 == 0) r--;
  cout << r << endl;
  return 0;
}

```