#include <bits/stdc++.h>
using namespace std;
long long n, t, data[55];
long long f[55][55], g[55][55];
int mem[55], top, last, D;
long long Ans;
int main() {
  scanf("%I64d%I64d", &n, &t);
  n++;
  data[0] = 1LL;
  for (int i = 1; i <= 40; i++) data[i] = data[i - 1] << 1LL;
  D = lower_bound(data, data + 41, t) - data;
  if (data[D] != t) {
    puts("0");
    return 0;
  }
  D++;
  f[1][1] = 1;
  g[1][1] = 1;
  g[0][0] = g[1][0] = 1;
  for (int i = 2; i <= 40; i++) {
    for (int j = 1; j <= i; j++) {
      f[i][j] = f[i - 1][j] + f[i - 1][j - 1];
      g[i][j] = g[i - 1][j] + f[i][j];
    }
    g[i][0] = 1;
  }
  for (top = 0; n; n >>= 1) mem[++top] = n & 1;
  int ck = 0;
  for (int i = 1; i <= top; i++)
    if (mem[i]) ck++;
  Ans = g[top - 1][D] + (ck == D);
  if (D == 1) Ans--;
  for (int i = top; i; i--) {
    int st = (i == top) ? (1) : (0);
    for (int j = st; j < mem[i]; j++) Ans += g[i - 1][D - last];
    if (mem[i]) last++;
    if (last > D) break;
  }
  printf("%I64d\n", Ans);
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
long long n, t, data[55];
long long f[55][55], g[55][55];
int mem[55], top, last, D;
long long Ans;
int main() {
  scanf("%I64d%I64d", &n, &t);
  n++;
  data[0] = 1LL;
  for (int i = 1; i <= 40; i++) data[i] = data[i - 1] << 1LL;
  D = lower_bound(data, data + 41, t) - data;
  if (data[D] != t) {
    puts("0");
    return 0;
  }
  D++;
  f[1][1] = 1;
  g[1][1] = 1;
  g[0][0] = g[1][0] = 1;
  for (int i = 2; i <= 40; i++) {
    for (int j = 1; j <= i; j++) {
      f[i][j] = f[i - 1][j] + f[i - 1][j - 1];
      g[i][j] = g[i - 1][j] + f[i][j];
    }
    g[i][0] = 1;
  }
  for (top = 0; n; n >>= 1) mem[++top] = n & 1;
  int ck = 0;
  for (int i = 1; i <= top; i++)
    if (mem[i]) ck++;
  Ans = g[top - 1][D] + (ck == D);
  if (D == 1) Ans--;
  for (int i = top; i; i--) {
    int st = (i == top) ? (1) : (0);
    for (int j = st; j < mem[i]; j++) Ans += g[i - 1][D - last];
    if (mem[i]) last++;
    if (last > D) break;
  }
  printf("%I64d\n", Ans);
  return 0;
}

```