#include <bits/stdc++.h>
using namespace std;
vector<int> bit;
long long mem[66][66][2];
long long n, t;
long long fact_power(long long a, long long b) {
  if (b < 0) return 0;
  long long res = 1;
  while (b) {
    if (b & 1) res *= a;
    a *= a;
    b >>= 1;
  }
  return res;
}
long long dp(int sum, int idx, int les) {
  if (idx == bit.size()) {
    return (fact_power(2, sum - 1) == t);
  }
  long long &ret = mem[sum][idx][les];
  if (ret != -1) {
    return ret;
  }
  ret = 0;
  if (les) {
    for (int i = 0; i < 2; i++) {
      ret += dp(sum + i, idx + 1, 1);
    }
  } else {
    for (int i = 0; i <= bit[idx]; i++) {
      ret += dp(sum + i, idx + 1, (i == bit[idx] ? 0 : 1));
    }
  }
  return ret;
}
int main() {
  cin >> n >> t;
  n++;
  while (n) {
    bit.push_back(n & 1);
    n >>= 1;
  }
  reverse(bit.begin(), bit.end());
  memset(mem, -1, sizeof(mem));
  if (t == 1) {
    cout << dp(0, 0, 0) - 1 << "\n";
  } else {
    cout << dp(0, 0, 0) << "\n";
  }
}

### 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;
vector<int> bit;
long long mem[66][66][2];
long long n, t;
long long fact_power(long long a, long long b) {
  if (b < 0) return 0;
  long long res = 1;
  while (b) {
    if (b & 1) res *= a;
    a *= a;
    b >>= 1;
  }
  return res;
}
long long dp(int sum, int idx, int les) {
  if (idx == bit.size()) {
    return (fact_power(2, sum - 1) == t);
  }
  long long &ret = mem[sum][idx][les];
  if (ret != -1) {
    return ret;
  }
  ret = 0;
  if (les) {
    for (int i = 0; i < 2; i++) {
      ret += dp(sum + i, idx + 1, 1);
    }
  } else {
    for (int i = 0; i <= bit[idx]; i++) {
      ret += dp(sum + i, idx + 1, (i == bit[idx] ? 0 : 1));
    }
  }
  return ret;
}
int main() {
  cin >> n >> t;
  n++;
  while (n) {
    bit.push_back(n & 1);
    n >>= 1;
  }
  reverse(bit.begin(), bit.end());
  memset(mem, -1, sizeof(mem));
  if (t == 1) {
    cout << dp(0, 0, 0) - 1 << "\n";
  } else {
    cout << dp(0, 0, 0) << "\n";
  }
}

```