#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
long long dp[55][55];
void init() {
  dp[0][0] = 1;
  for (int i = 1; i <= 52; i++)
    for (int j = 0; j <= i; j++) {
      dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1];
    }
}
long long two(long long i) { return (long long)1 << i; }
int p;
void solve(long long n) {
  n++;
  int dig[55], tot = 0;
  int pos = -1;
  for (int i = 52; i >= 0; i--)
    if (two(i) & n) {
      pos = i;
      break;
    }
  for (int i = pos; i >= 0; i--)
    if (two(i) & n)
      dig[tot++] = 1;
    else
      dig[tot++] = 0;
  long long ans = 0;
  long long cnt = 0;
  for (int i = 0; i < tot; i++)
    if (dig[i]) {
      if (p >= cnt) ans += dp[tot - i - 1][p - cnt];
      cnt++;
    }
  if (cnt == p) ans++;
  if (p == 1) ans--;
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(false);
  init();
  long long n, t;
  cin >> n >> t;
  int flag = 0;
  for (int i = 0; i < 52; i++) {
    if (two(i) == t) {
      flag = 1;
      p = i;
    }
  }
  p++;
  if (!flag) {
    puts("0");
    return 0;
  }
  solve(n);
}

### 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>
using namespace std;
const int inf = 0x3f3f3f3f;
long long dp[55][55];
void init() {
  dp[0][0] = 1;
  for (int i = 1; i <= 52; i++)
    for (int j = 0; j <= i; j++) {
      dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1];
    }
}
long long two(long long i) { return (long long)1 << i; }
int p;
void solve(long long n) {
  n++;
  int dig[55], tot = 0;
  int pos = -1;
  for (int i = 52; i >= 0; i--)
    if (two(i) & n) {
      pos = i;
      break;
    }
  for (int i = pos; i >= 0; i--)
    if (two(i) & n)
      dig[tot++] = 1;
    else
      dig[tot++] = 0;
  long long ans = 0;
  long long cnt = 0;
  for (int i = 0; i < tot; i++)
    if (dig[i]) {
      if (p >= cnt) ans += dp[tot - i - 1][p - cnt];
      cnt++;
    }
  if (cnt == p) ans++;
  if (p == 1) ans--;
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(false);
  init();
  long long n, t;
  cin >> n >> t;
  int flag = 0;
  for (int i = 0; i < 52; i++) {
    if (two(i) == t) {
      flag = 1;
      p = i;
    }
  }
  p++;
  if (!flag) {
    puts("0");
    return 0;
  }
  solve(n);
}

```