#include <bits/stdc++.h>
using namespace std;
long long n, po;
long long c[50][50];
long long ans;
inline void init() {
  ans = 0;
  memset(c, 0, sizeof(c));
  c[0][0] = 1;
  for (register long long i = 1; i < 50; ++i) {
    c[i][0] = 1;
    for (register long long j = 1; j < 50; ++j)
      c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
  }
  return;
}
inline long long dfs(long long pos, long long len, long long cnt,
                     long long cot) {
  if (cot < 0 || cot > cnt) return 0;
  if (len == 1) return 1;
  long long mid = len / 2;
  if (pos > mid)
    return c[cnt - 1][cot] + dfs(pos - mid, mid, cnt - 1, cot - 1);
  else
    return dfs(pos, mid, cnt - 1, cot);
}
int main() {
  init();
  scanf("%I64d%I64d", &n, &po);
  long long x = 1, cot = 0;
  while (x < po) {
    x <<= 1;
    cot++;
  }
  if (x != po) {
    printf("0\n");
    return 0;
  }
  x = 2;
  long long y = 0, cnt = 0;
  while (y < n) {
    y += x;
    x <<= 1;
    cnt++;
  }
  long long xx = cot > 1 ? cot : 1;
  for (register long long i = xx; i < cnt; ++i) ans += c[i][cot];
  ans += dfs(n - (y - x / 2), x / 2, cnt, cot);
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Please create a solution in CPP 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;
long long n, po;
long long c[50][50];
long long ans;
inline void init() {
  ans = 0;
  memset(c, 0, sizeof(c));
  c[0][0] = 1;
  for (register long long i = 1; i < 50; ++i) {
    c[i][0] = 1;
    for (register long long j = 1; j < 50; ++j)
      c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
  }
  return;
}
inline long long dfs(long long pos, long long len, long long cnt,
                     long long cot) {
  if (cot < 0 || cot > cnt) return 0;
  if (len == 1) return 1;
  long long mid = len / 2;
  if (pos > mid)
    return c[cnt - 1][cot] + dfs(pos - mid, mid, cnt - 1, cot - 1);
  else
    return dfs(pos, mid, cnt - 1, cot);
}
int main() {
  init();
  scanf("%I64d%I64d", &n, &po);
  long long x = 1, cot = 0;
  while (x < po) {
    x <<= 1;
    cot++;
  }
  if (x != po) {
    printf("0\n");
    return 0;
  }
  x = 2;
  long long y = 0, cnt = 0;
  while (y < n) {
    y += x;
    x <<= 1;
    cnt++;
  }
  long long xx = cot > 1 ? cot : 1;
  for (register long long i = xx; i < cnt; ++i) ans += c[i][cot];
  ans += dfs(n - (y - x / 2), x / 2, cnt, cot);
  printf("%I64d\n", ans);
  return 0;
}

```