#include <bits/stdc++.h>
using namespace std;
long long n, t;
long long g(long long n, long long t);
long long f(long long type, long long depth, long long t);
long long f(long long type, long long depth, long long t) {
  if (depth <= 0) return 0;
  if (t == 1) return 1;
  if (type == 1) return (t == 1) ? 1 : 0;
  if (t % 2) return 0;
  return g(min(type - 1, depth - 1), t / 2);
}
map<pair<long long, long long>, long long> memo;
long long g(long long n, long long t) {
  map<pair<long long, long long>, long long>::iterator it =
      memo.find(pair<long long, long long>(n, t));
  if (it != memo.end()) return it->second;
  long long res = 0;
  for (long long type = 1; type <= n; type *= 2) {
    res += f(type, min(n - type + 1, type), t);
  }
  return memo[pair<long long, long long>(n, t)] = res;
}
int main() {
  scanf("%I64d %I64d ", &n, &t);
  long long res = g(n + 1, t);
  if (t == 1) res--;
  printf("%I64d\n", res);
}

### Prompt

In cpp, your task is to solve 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;
long long n, t;
long long g(long long n, long long t);
long long f(long long type, long long depth, long long t);
long long f(long long type, long long depth, long long t) {
  if (depth <= 0) return 0;
  if (t == 1) return 1;
  if (type == 1) return (t == 1) ? 1 : 0;
  if (t % 2) return 0;
  return g(min(type - 1, depth - 1), t / 2);
}
map<pair<long long, long long>, long long> memo;
long long g(long long n, long long t) {
  map<pair<long long, long long>, long long>::iterator it =
      memo.find(pair<long long, long long>(n, t));
  if (it != memo.end()) return it->second;
  long long res = 0;
  for (long long type = 1; type <= n; type *= 2) {
    res += f(type, min(n - type + 1, type), t);
  }
  return memo[pair<long long, long long>(n, t)] = res;
}
int main() {
  scanf("%I64d %I64d ", &n, &t);
  long long res = g(n + 1, t);
  if (t == 1) res--;
  printf("%I64d\n", res);
}

```