#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:66777216")
using namespace std;
long long n, t, dp[2][60][60], C[60][60];
void calcBinomials(int n = 59) {
  int i, j;
  for (i = 0; i <= n; ++i) {
    C[i][0] = 1;
  }
  for (i = 1; i <= n; ++i) {
    for (j = 1; j <= n; ++j) {
      C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
    }
  }
}
int main() {
  calcBinomials();
  cin >> n >> t;
  ++n;
  int x = -1;
  for (int i = 0; i < 50; ++i) {
    if ((1LL << i) == t) {
      x = i;
    }
  }
  if (x == -1) {
    puts("0");
    return 0;
  }
  int nlen = 1;
  while ((1LL << nlen) <= n) {
    nlen++;
  }
  int dig[60];
  for (int i = 0; i < nlen; ++i) {
    dig[nlen - i] = ((n >> i) & 1);
  }
  long long ans = 0LL;
  for (int len = 1; len < nlen; ++len) {
    int ones = x + 1;
    if (ones <= 0 || ones > len) {
      continue;
    }
    ans += C[len - 1][ones - 1];
  }
  dp[1][0][0] = 1LL;
  for (int len = 1; len <= nlen; ++len) {
    for (int ones = 1; ones <= len; ++ones) {
      dp[1][len][ones] = dp[1][len - 1][ones - dig[len]];
      dp[0][len][ones] = dp[0][len - 1][ones - 1] + dp[0][len - 1][ones];
      if (dig[len] == 1) {
        dp[0][len][ones] += dp[1][len - 1][ones];
      }
    }
  }
  ans += dp[1][nlen][x + 1] + dp[0][nlen][x + 1];
  if (x == 0) {
    --ans;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please formulate 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>
#pragma comment(linker, "/STACK:66777216")
using namespace std;
long long n, t, dp[2][60][60], C[60][60];
void calcBinomials(int n = 59) {
  int i, j;
  for (i = 0; i <= n; ++i) {
    C[i][0] = 1;
  }
  for (i = 1; i <= n; ++i) {
    for (j = 1; j <= n; ++j) {
      C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
    }
  }
}
int main() {
  calcBinomials();
  cin >> n >> t;
  ++n;
  int x = -1;
  for (int i = 0; i < 50; ++i) {
    if ((1LL << i) == t) {
      x = i;
    }
  }
  if (x == -1) {
    puts("0");
    return 0;
  }
  int nlen = 1;
  while ((1LL << nlen) <= n) {
    nlen++;
  }
  int dig[60];
  for (int i = 0; i < nlen; ++i) {
    dig[nlen - i] = ((n >> i) & 1);
  }
  long long ans = 0LL;
  for (int len = 1; len < nlen; ++len) {
    int ones = x + 1;
    if (ones <= 0 || ones > len) {
      continue;
    }
    ans += C[len - 1][ones - 1];
  }
  dp[1][0][0] = 1LL;
  for (int len = 1; len <= nlen; ++len) {
    for (int ones = 1; ones <= len; ++ones) {
      dp[1][len][ones] = dp[1][len - 1][ones - dig[len]];
      dp[0][len][ones] = dp[0][len - 1][ones - 1] + dp[0][len - 1][ones];
      if (dig[len] == 1) {
        dp[0][len][ones] += dp[1][len - 1][ones];
      }
    }
  }
  ans += dp[1][nlen][x + 1] + dp[0][nlen][x + 1];
  if (x == 0) {
    --ans;
  }
  cout << ans << endl;
  return 0;
}

```