#include <bits/stdc++.h>
using namespace std;
long long int n, t, N, k;
int p[64];
long long int PD[64][64][2];
long long int calc(int u, int c, int l) {
  if (u == 50) {
    if (c == k + 1) return 1ll;
    return 0;
  }
  long long int &pd = PD[u][c][l];
  if (pd != -1) return pd;
  long long int soma = 0;
  soma += calc(u + 1, c, (l ? 1 : p[u]));
  if (l or p[u]) soma += calc(u + 1, c + 1, l);
  return pd = soma;
}
int main() {
  memset(PD, 0xff, sizeof(PD));
  cin >> n >> t;
  if (__builtin_popcount(t) != 1) {
    printf("0\n");
    return 0;
  }
  for (k = 0;; k++)
    if ((1ll << k) == t) break;
  N = n + 1;
  for (int i = 0; i < 50; i++) p[i] = (N & (1ll << ((50 - 1) - i))) ? 1 : 0;
  long long int tot = calc(0, 0, 0);
  if (__builtin_popcount(0) == k + 1) tot--;
  if (__builtin_popcount(1) == k + 1) tot--;
  cout << tot << endl;
  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;
long long int n, t, N, k;
int p[64];
long long int PD[64][64][2];
long long int calc(int u, int c, int l) {
  if (u == 50) {
    if (c == k + 1) return 1ll;
    return 0;
  }
  long long int &pd = PD[u][c][l];
  if (pd != -1) return pd;
  long long int soma = 0;
  soma += calc(u + 1, c, (l ? 1 : p[u]));
  if (l or p[u]) soma += calc(u + 1, c + 1, l);
  return pd = soma;
}
int main() {
  memset(PD, 0xff, sizeof(PD));
  cin >> n >> t;
  if (__builtin_popcount(t) != 1) {
    printf("0\n");
    return 0;
  }
  for (k = 0;; k++)
    if ((1ll << k) == t) break;
  N = n + 1;
  for (int i = 0; i < 50; i++) p[i] = (N & (1ll << ((50 - 1) - i))) ? 1 : 0;
  long long int tot = calc(0, 0, 0);
  if (__builtin_popcount(0) == k + 1) tot--;
  if (__builtin_popcount(1) == k + 1) tot--;
  cout << tot << endl;
  return 0;
}

```