#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000100;
const int INF = 1e9 + 10;
long long n, t;
int goal;
long long f[60][60];
int num[maxn], len;
long long dfs(int i, int one, bool e) {
  if (i == -1) return one == goal;
  if (!e && ~f[i][one]) return f[i][one];
  long long res = 0;
  int u = e ? num[i] : 1;
  for (int d = 0; d <= u; d++) {
    res += dfs(i - 1, one + d, e && d == u);
  }
  return e ? res : f[i][one] = res;
}
long long F(long long n) {
  len = 0;
  while (n) {
    num[len++] = n % 2;
    n >>= 1;
  }
  return dfs(len - 1, 0, 1);
}
bool ispow2(long long n) {
  int cnt = 0;
  while (n) {
    cnt += n % 2;
    n >>= 1;
  }
  return cnt == 1;
}
int main() {
  while (cin >> n >> t) {
    memset(f, -1, sizeof(f));
    if (!ispow2(t))
      puts("0");
    else {
      goal = 0;
      while (t) goal++, t >>= 1;
      printf("%I64d\n", F(n + 1) - F(1));
    }
  }
  return 0;
}

### 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;
const int maxn = 1000100;
const int INF = 1e9 + 10;
long long n, t;
int goal;
long long f[60][60];
int num[maxn], len;
long long dfs(int i, int one, bool e) {
  if (i == -1) return one == goal;
  if (!e && ~f[i][one]) return f[i][one];
  long long res = 0;
  int u = e ? num[i] : 1;
  for (int d = 0; d <= u; d++) {
    res += dfs(i - 1, one + d, e && d == u);
  }
  return e ? res : f[i][one] = res;
}
long long F(long long n) {
  len = 0;
  while (n) {
    num[len++] = n % 2;
    n >>= 1;
  }
  return dfs(len - 1, 0, 1);
}
bool ispow2(long long n) {
  int cnt = 0;
  while (n) {
    cnt += n % 2;
    n >>= 1;
  }
  return cnt == 1;
}
int main() {
  while (cin >> n >> t) {
    memset(f, -1, sizeof(f));
    if (!ispow2(t))
      puts("0");
    else {
      goal = 0;
      while (t) goal++, t >>= 1;
      printf("%I64d\n", F(n + 1) - F(1));
    }
  }
  return 0;
}

```