#include <bits/stdc++.h>
using namespace std;
map<long long, long long> mp[64];
long long dfs(int x, long long y) {
  long long tmp;
  if (mp[x].count(y)) return mp[x][y];
  if (x == 0) {
    if (y == 1) {
      return mp[x][y] = 1;
    } else {
      return mp[x][y] = 0;
    }
  }
  tmp = dfs(x - 1, y);
  if (y % 2 == 0) tmp += dfs(x - 1, y / 2);
  return mp[x][y] = tmp;
}
int main() {
  long long n, t, m, sum = 0, i, j;
  scanf("%I64d %I64d", &n, &t);
  n++;
  m = n;
  for (i = 0; m > 0; i++) m >>= 1;
  for (j = 0; j < i - 1; j++) sum += dfs(j, t);
  n -= (1LL << (i - 1)) - 1;
  if (t == 1) sum--;
  while (n) {
    m = n;
    for (i = 0; m > 0; i++) m >>= 1;
    sum += dfs(i - 1, t);
    n -= (1LL << (i - 1));
    if (t % 2 != 0) break;
    t /= 2;
  }
  printf("%I64d\n", sum);
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
map<long long, long long> mp[64];
long long dfs(int x, long long y) {
  long long tmp;
  if (mp[x].count(y)) return mp[x][y];
  if (x == 0) {
    if (y == 1) {
      return mp[x][y] = 1;
    } else {
      return mp[x][y] = 0;
    }
  }
  tmp = dfs(x - 1, y);
  if (y % 2 == 0) tmp += dfs(x - 1, y / 2);
  return mp[x][y] = tmp;
}
int main() {
  long long n, t, m, sum = 0, i, j;
  scanf("%I64d %I64d", &n, &t);
  n++;
  m = n;
  for (i = 0; m > 0; i++) m >>= 1;
  for (j = 0; j < i - 1; j++) sum += dfs(j, t);
  n -= (1LL << (i - 1)) - 1;
  if (t == 1) sum--;
  while (n) {
    m = n;
    for (i = 0; m > 0; i++) m >>= 1;
    sum += dfs(i - 1, t);
    n -= (1LL << (i - 1));
    if (t % 2 != 0) break;
    t /= 2;
  }
  printf("%I64d\n", sum);
  return 0;
}

```