#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:64000000")
const int maxn = 1 << 10, maxm = 60;
const int inf = 1000000007;
const int mod = 1000000007;
long long n, t;
int a[maxn][maxn];
long long c[maxm][maxm];
void precalc() {
  c[0][0] = 1;
  for (int i = 0; i < maxm - 1; i++)
    for (int j = 0; j <= i; j++) {
      c[i + 1][j] += c[i][j];
      c[i + 1][j + 1] += c[i][j];
    }
}
void build(int m) {
  for (int i = 0; i < m; i++) a[0][i] = 0;
  a[0][m] = 1;
  for (int i = 1; i <= m; i++) {
    for (int j = 0; j <= m; j++) {
      if (j == 0) {
        a[i][j] = a[i - 1][j + 1];
      } else {
        if (j == m) {
          a[i][j] = a[i - 1][j - 1];
        } else {
          a[i][j] = a[i - 1][j - 1] ^ a[i - 1][j + 1];
        }
      }
    }
  }
}
void print(int m) {
  for (int i = 0; i <= m; i++)
    for (int j = 0; j <= m; j++) cout << a[i][j] << " \n"[j == m];
}
int sum(int m) {
  int res = 0;
  for (int i = 0; i <= m; i++) res += a[m][i];
  return res;
}
int lg(long long t) {
  int a = 0;
  while (t) {
    t >>= 1;
    a++;
  }
  return a;
}
long long calc(long long x, int y) {
  if (y == 0) return 1;
  if (x == 0) return y == 0;
  long long res = c[lg(x) - 1][y];
  res += calc(x - (1LL << (lg(x) - 1)), y - 1);
  return res;
}
int main() {
  precalc();
  cin >> n >> t;
  if (t & (t - 1)) {
    cout << 0;
    return 0;
  }
  int a = lg(t);
  cout << calc(n + 1, a) - (t == 1);
  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;
#pragma comment(linker, "/STACK:64000000")
const int maxn = 1 << 10, maxm = 60;
const int inf = 1000000007;
const int mod = 1000000007;
long long n, t;
int a[maxn][maxn];
long long c[maxm][maxm];
void precalc() {
  c[0][0] = 1;
  for (int i = 0; i < maxm - 1; i++)
    for (int j = 0; j <= i; j++) {
      c[i + 1][j] += c[i][j];
      c[i + 1][j + 1] += c[i][j];
    }
}
void build(int m) {
  for (int i = 0; i < m; i++) a[0][i] = 0;
  a[0][m] = 1;
  for (int i = 1; i <= m; i++) {
    for (int j = 0; j <= m; j++) {
      if (j == 0) {
        a[i][j] = a[i - 1][j + 1];
      } else {
        if (j == m) {
          a[i][j] = a[i - 1][j - 1];
        } else {
          a[i][j] = a[i - 1][j - 1] ^ a[i - 1][j + 1];
        }
      }
    }
  }
}
void print(int m) {
  for (int i = 0; i <= m; i++)
    for (int j = 0; j <= m; j++) cout << a[i][j] << " \n"[j == m];
}
int sum(int m) {
  int res = 0;
  for (int i = 0; i <= m; i++) res += a[m][i];
  return res;
}
int lg(long long t) {
  int a = 0;
  while (t) {
    t >>= 1;
    a++;
  }
  return a;
}
long long calc(long long x, int y) {
  if (y == 0) return 1;
  if (x == 0) return y == 0;
  long long res = c[lg(x) - 1][y];
  res += calc(x - (1LL << (lg(x) - 1)), y - 1);
  return res;
}
int main() {
  precalc();
  cin >> n >> t;
  if (t & (t - 1)) {
    cout << 0;
    return 0;
  }
  int a = lg(t);
  cout << calc(n + 1, a) - (t == 1);
  return 0;
}

```