#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:64000000")
using namespace std;
long long c[100][100];
long long len[100];
long long calc(int k, long long n, int pw) {
  if (k == 1) {
    if (n == 0) return 0;
    if (n == 1) {
      return pw == 0 ? 1 : 0;
    }
    return c[k][pw];
  }
  if (n < len[k - 1]) return calc(k - 1, n, pw);
  long long ret = c[k - 1][pw];
  n -= len[k - 1];
  if (pw > 0) ret += calc(k - 1, n, pw - 1);
  return ret;
}
int main() {
  long long n, t;
  c[1][0] = 1;
  c[1][1] = 1;
  len[1] = 2;
  for (int k = 2; k < 63; k++) {
    c[k][0] = c[k - 1][0];
    for (int i = 1; i <= k; i++) {
      c[k][i] = c[k - 1][i] + c[k - 1][i - 1];
    }
    len[k] = len[k - 1] * 2;
  }
  cin >> n >> t;
  long long nn = n;
  int pw = -1;
  for (int i = 0; i < (int)(60); i++)
    if (t == (1ll << i)) pw = i;
  if (pw == -1) {
    cout << 0 << endl;
    return 0;
  }
  int kk = 1;
  long long ans = 0;
  while (len[kk] <= n) {
    ans += c[kk][pw];
    n -= len[kk];
    kk++;
  }
  ans += calc(kk, n, pw);
  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:64000000")
using namespace std;
long long c[100][100];
long long len[100];
long long calc(int k, long long n, int pw) {
  if (k == 1) {
    if (n == 0) return 0;
    if (n == 1) {
      return pw == 0 ? 1 : 0;
    }
    return c[k][pw];
  }
  if (n < len[k - 1]) return calc(k - 1, n, pw);
  long long ret = c[k - 1][pw];
  n -= len[k - 1];
  if (pw > 0) ret += calc(k - 1, n, pw - 1);
  return ret;
}
int main() {
  long long n, t;
  c[1][0] = 1;
  c[1][1] = 1;
  len[1] = 2;
  for (int k = 2; k < 63; k++) {
    c[k][0] = c[k - 1][0];
    for (int i = 1; i <= k; i++) {
      c[k][i] = c[k - 1][i] + c[k - 1][i - 1];
    }
    len[k] = len[k - 1] * 2;
  }
  cin >> n >> t;
  long long nn = n;
  int pw = -1;
  for (int i = 0; i < (int)(60); i++)
    if (t == (1ll << i)) pw = i;
  if (pw == -1) {
    cout << 0 << endl;
    return 0;
  }
  int kk = 1;
  long long ans = 0;
  while (len[kk] <= n) {
    ans += c[kk][pw];
    n -= len[kk];
    kk++;
  }
  ans += calc(kk, n, pw);
  cout << ans << endl;
  return 0;
}

```