#include <bits/stdc++.h>
using namespace std;
ifstream fin("in.in");
ofstream fout("out.out");
const int N = 43;
unsigned long long n, t, dp[N][N], a[N], res;
int len;
bool check() {
  unsigned long long tmp = t;
  while (tmp > 1) {
    if (tmp & 1) return false;
    tmp >>= 1;
    len++;
  }
  if (len > 42) return false;
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  dp[0][0] = 1;
  for (int l = 1; l < N; l++) {
    dp[l][0] = 1;
    for (int i = 1; i < N; i++) dp[l][i] = dp[l - 1][i] + dp[l - 1][i - 1];
  }
  a[0] = 1;
  for (int i = 1; i < N; i++) a[i] = a[i - 1] << 1;
  cin >> n >> t;
  if (n == 1) {
    if (t == 1)
      cout << 1;
    else
      cout << 0;
    return 0;
  }
  n++;
  if (t == 1) {
    while (n) {
      n >>= 1;
      res++;
    }
    cout << res - 1;
    return 0;
  }
  if (!check()) return (cout << 0), 0;
  int l;
  for (l = 0; a[l] <= n; l++) {
    res += dp[l][len];
    n -= a[l];
  }
  while (n && len >= 0) {
    if (n == a[l]) {
      res += dp[l][len];
      n = 0;
      continue;
    }
    if (n <= a[l] / 2) {
      l--;
      continue;
    }
    res += dp[l - 1][len];
    n -= a[l - 1];
    l--;
    len--;
  }
  cout << res << endl;
  return 0;
}

### Prompt

Your task is to create 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;
ifstream fin("in.in");
ofstream fout("out.out");
const int N = 43;
unsigned long long n, t, dp[N][N], a[N], res;
int len;
bool check() {
  unsigned long long tmp = t;
  while (tmp > 1) {
    if (tmp & 1) return false;
    tmp >>= 1;
    len++;
  }
  if (len > 42) return false;
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  dp[0][0] = 1;
  for (int l = 1; l < N; l++) {
    dp[l][0] = 1;
    for (int i = 1; i < N; i++) dp[l][i] = dp[l - 1][i] + dp[l - 1][i - 1];
  }
  a[0] = 1;
  for (int i = 1; i < N; i++) a[i] = a[i - 1] << 1;
  cin >> n >> t;
  if (n == 1) {
    if (t == 1)
      cout << 1;
    else
      cout << 0;
    return 0;
  }
  n++;
  if (t == 1) {
    while (n) {
      n >>= 1;
      res++;
    }
    cout << res - 1;
    return 0;
  }
  if (!check()) return (cout << 0), 0;
  int l;
  for (l = 0; a[l] <= n; l++) {
    res += dp[l][len];
    n -= a[l];
  }
  while (n && len >= 0) {
    if (n == a[l]) {
      res += dp[l][len];
      n = 0;
      continue;
    }
    if (n <= a[l] / 2) {
      l--;
      continue;
    }
    res += dp[l - 1][len];
    n -= a[l - 1];
    l--;
    len--;
  }
  cout << res << endl;
  return 0;
}

```