#include <bits/stdc++.h>
using namespace std;
const int maxt = 45;
long long C[maxt][maxt];
void solve(long long n, int cnt) {
  int cntn = 0;
  for (long long x = n; x; x >>= 1, cntn++)
    ;
  assert(cntn < maxt);
  long long ans = 0;
  int need = cnt, left = cntn - 1;
  for (int put = 0; put <= cntn && need >= 0; put++) {
    if (put == cntn || ((n >> (cntn - 1 - put)) & 1))
      ans += C[max(0, left)][need];
    if (put == cntn) break;
    need -= ((n >> (cntn - put - 1)) & 1);
    left--;
  }
  if (cnt == 1) ans--;
  printf(
      "%lld"
      "\n",
      ans);
}
int main() {
  {
    memset(C, 0, sizeof(C));
    for (int n = 0; n < maxt; n++) {
      C[n][0] = C[n][n] = 1;
      for (int k = 1; k < n; k++) C[n][k] = C[n - 1][k] + C[n - 1][k - 1];
    }
  }
  long long n, t;
  while (scanf("%lld"
               "%lld",
               &n, &t) >= 1) {
    if (t & (t - 1)) {
      printf("0\n");
      continue;
    }
    int cnt = 0;
    for (; t; t >>= 1, cnt++)
      ;
    solve(n + 1, cnt);
  }
  return 0;
}

### 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;
const int maxt = 45;
long long C[maxt][maxt];
void solve(long long n, int cnt) {
  int cntn = 0;
  for (long long x = n; x; x >>= 1, cntn++)
    ;
  assert(cntn < maxt);
  long long ans = 0;
  int need = cnt, left = cntn - 1;
  for (int put = 0; put <= cntn && need >= 0; put++) {
    if (put == cntn || ((n >> (cntn - 1 - put)) & 1))
      ans += C[max(0, left)][need];
    if (put == cntn) break;
    need -= ((n >> (cntn - put - 1)) & 1);
    left--;
  }
  if (cnt == 1) ans--;
  printf(
      "%lld"
      "\n",
      ans);
}
int main() {
  {
    memset(C, 0, sizeof(C));
    for (int n = 0; n < maxt; n++) {
      C[n][0] = C[n][n] = 1;
      for (int k = 1; k < n; k++) C[n][k] = C[n - 1][k] + C[n - 1][k - 1];
    }
  }
  long long n, t;
  while (scanf("%lld"
               "%lld",
               &n, &t) >= 1) {
    if (t & (t - 1)) {
      printf("0\n");
      continue;
    }
    int cnt = 0;
    for (; t; t >>= 1, cnt++)
      ;
    solve(n + 1, cnt);
  }
  return 0;
}

```