#include <bits/stdc++.h>
using namespace std;
long long dp[70][200];
long long t;
long long two[70], two_sum[70];
map<pair<long long, long long>, long long> memo;
long long go(long long n, long long pw) {
  if (n == 0) return 0;
  if (n == 1) return pw == t;
  if (memo.count(make_pair(n, pw))) return memo[make_pair(n, pw)];
  long long &ret = memo[make_pair(n, pw)];
  int idx = 0;
  for (int i = 60; i >= 0; --i)
    if (two_sum[i] <= n) {
      idx = i;
      break;
    }
  if (two_sum[idx] == n) {
    ret = go(two_sum[idx - 1], pw) + go(two_sum[idx - 1], pw) -
          go(two_sum[idx - 2], pw) + go(two_sum[idx - 1], pw * 2) -
          go(two_sum[idx - 2], pw * 2);
  } else {
    ret = go(two_sum[idx], pw);
    long long rest = n - two_sum[idx];
    if (rest > two[idx]) {
      ret += go(two_sum[idx], pw) - go(two_sum[idx - 1], pw);
      ret += go(two_sum[idx - 1] + rest - two[idx], pw * 2) -
             go(two_sum[idx - 1], pw * 2);
    } else {
      ret += go(two_sum[idx - 1] + rest, pw) - go(two_sum[idx - 1], pw);
    }
  }
  return ret;
}
int main() {
  memset(dp, -1, sizeof(dp));
  two[0] = two_sum[0] = 1;
  for (int i = 1; i <= 60; ++i)
    two[i] = two[i - 1] * 2, two_sum[i] = two_sum[i - 1] + two[i];
  long long n;
  cin >> n >> t;
  cout << go(n + 1, 1) - (t == 1);
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
long long dp[70][200];
long long t;
long long two[70], two_sum[70];
map<pair<long long, long long>, long long> memo;
long long go(long long n, long long pw) {
  if (n == 0) return 0;
  if (n == 1) return pw == t;
  if (memo.count(make_pair(n, pw))) return memo[make_pair(n, pw)];
  long long &ret = memo[make_pair(n, pw)];
  int idx = 0;
  for (int i = 60; i >= 0; --i)
    if (two_sum[i] <= n) {
      idx = i;
      break;
    }
  if (two_sum[idx] == n) {
    ret = go(two_sum[idx - 1], pw) + go(two_sum[idx - 1], pw) -
          go(two_sum[idx - 2], pw) + go(two_sum[idx - 1], pw * 2) -
          go(two_sum[idx - 2], pw * 2);
  } else {
    ret = go(two_sum[idx], pw);
    long long rest = n - two_sum[idx];
    if (rest > two[idx]) {
      ret += go(two_sum[idx], pw) - go(two_sum[idx - 1], pw);
      ret += go(two_sum[idx - 1] + rest - two[idx], pw * 2) -
             go(two_sum[idx - 1], pw * 2);
    } else {
      ret += go(two_sum[idx - 1] + rest, pw) - go(two_sum[idx - 1], pw);
    }
  }
  return ret;
}
int main() {
  memset(dp, -1, sizeof(dp));
  two[0] = two_sum[0] = 1;
  for (int i = 1; i <= 60; ++i)
    two[i] = two[i - 1] * 2, two_sum[i] = two_sum[i - 1] + two[i];
  long long n;
  cin >> n >> t;
  cout << go(n + 1, 1) - (t == 1);
}

```