#include <bits/stdc++.h>
using namespace std;
const double eps(1e-8);
long long dp[60][60];
int main() {
  memset(dp, 0, sizeof(dp));
  dp[0][0] = 1;
  for (int i = 1; i < 60; i++) {
    dp[i][0] = 1;
    for (int j = 1; j < 60; j++) dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
  }
  long long n, t;
  scanf("%I64d %I64d", &n, &t);
  n++;
  if (__builtin_popcount(t) != 1) {
    printf("0\n");
    return 0;
  }
  int bit = 0;
  while ((1LL << bit) != t) bit++;
  int now = 0;
  long long ans = 0;
  while ((1LL << now) <= n) {
    ans += dp[now][bit];
    n -= (1LL << now);
    now++;
  }
  while (n) {
    now--;
    if ((1LL << now) <= n) {
      n -= (1LL << now);
      ans += dp[now][bit];
      bit--;
    }
  }
  if (t == 1) ans--;
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 double eps(1e-8);
long long dp[60][60];
int main() {
  memset(dp, 0, sizeof(dp));
  dp[0][0] = 1;
  for (int i = 1; i < 60; i++) {
    dp[i][0] = 1;
    for (int j = 1; j < 60; j++) dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
  }
  long long n, t;
  scanf("%I64d %I64d", &n, &t);
  n++;
  if (__builtin_popcount(t) != 1) {
    printf("0\n");
    return 0;
  }
  int bit = 0;
  while ((1LL << bit) != t) bit++;
  int now = 0;
  long long ans = 0;
  while ((1LL << now) <= n) {
    ans += dp[now][bit];
    n -= (1LL << now);
    now++;
  }
  while (n) {
    now--;
    if ((1LL << now) <= n) {
      n -= (1LL << now);
      ans += dp[now][bit];
      bit--;
    }
  }
  if (t == 1) ans--;
  cout << ans << endl;
  return 0;
}

```