#include <bits/stdc++.h>
using namespace std;
long long memo[2][65][65];
int cnt, bcn;
bool mark[2][65][65];
char arr[65];
int getBinary(long long n) {
  if (n == 1) {
    arr[0] = '1';
    return 1;
  }
  int aux = getBinary(n / 2);
  arr[aux] = (n % 2) + '0';
  return aux + 1;
}
long long DP(int gt, int bits, int on) {
  if (bits >= bcn && on - 1 == cnt) return 1;
  if (bits >= bcn || on - 1 > cnt) return 0;
  long long &best = memo[gt][bits][on];
  if (mark[gt][bits][on]) return best;
  mark[gt][bits][on] = true;
  if (arr[bits] == '1')
    best = DP(true, bits + 1, on) + DP(gt, bits + 1, on + 1);
  else if (gt)
    best = (DP(gt, bits + 1, on + 1) + DP(gt, bits + 1, on));
  else
    best = DP(gt, bits + 1, on);
  return best;
}
int main() {
  long long n, t;
  scanf("%I64d %I64d", &n, &t);
  ++n;
  long long pt = 1LL;
  while (pt < t) pt <<= 1LL, cnt++;
  if (pt != t) {
    printf("0\n");
    return 0;
  }
  bcn = getBinary(n);
  printf("%I64d\n", DP(0, 0, 0) - (t == 1));
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 memo[2][65][65];
int cnt, bcn;
bool mark[2][65][65];
char arr[65];
int getBinary(long long n) {
  if (n == 1) {
    arr[0] = '1';
    return 1;
  }
  int aux = getBinary(n / 2);
  arr[aux] = (n % 2) + '0';
  return aux + 1;
}
long long DP(int gt, int bits, int on) {
  if (bits >= bcn && on - 1 == cnt) return 1;
  if (bits >= bcn || on - 1 > cnt) return 0;
  long long &best = memo[gt][bits][on];
  if (mark[gt][bits][on]) return best;
  mark[gt][bits][on] = true;
  if (arr[bits] == '1')
    best = DP(true, bits + 1, on) + DP(gt, bits + 1, on + 1);
  else if (gt)
    best = (DP(gt, bits + 1, on + 1) + DP(gt, bits + 1, on));
  else
    best = DP(gt, bits + 1, on);
  return best;
}
int main() {
  long long n, t;
  scanf("%I64d %I64d", &n, &t);
  ++n;
  long long pt = 1LL;
  while (pt < t) pt <<= 1LL, cnt++;
  if (pt != t) {
    printf("0\n");
    return 0;
  }
  bcn = getBinary(n);
  printf("%I64d\n", DP(0, 0, 0) - (t == 1));
  return 0;
}

```