#include <bits/stdc++.h>
using namespace std;
int a[560][560] = {0};
int build(int m, bool out = true) {
  memset(a, 0, sizeof(a));
  for (int j = 1; j <= m; j++) a[1][j] = 0;
  a[1][m + 1] = 1;
  for (int i = 2; i <= m + 1; ++i)
    for (int j = 1; j <= m + 1; ++j)
      if (j == 1)
        a[i][j] = a[i - 1][j + 1];
      else {
        if (j == m + 1)
          a[i][j] = a[i - 1][j - 1];
        else
          a[i][j] = a[i - 1][j - 1] ^ a[i - 1][j + 1];
      }
  int ret = 0;
  if (out) cout << "For:	" << m << endl;
  for (int i = 1; i <= m + 1; ++i) {
    ret += a[m + 1][i];
    for (int j = 1; j <= m + 1; ++j)
      if (out) cout << a[i][j] << "	";
    if (out) cout << endl;
  }
  if (out) cout << endl;
  return ret;
}
int C(int n, int k) {
  int res = 0;
  for (int i = n - k + 1; i <= n; ++i) {
    int j = i;
    while (j % 2 == 0) ++res, j /= 2;
  }
  for (int i = 1; i <= k; ++i) {
    int j = i;
    while (j % 2 == 0) --res, j /= 2;
  }
  return !res;
}
int bit(int x) {
  int ret = 0;
  while (x) {
    ret += (x & 1);
    x >>= 1;
  }
  return ret;
}
vector<int> v;
long long dp[129][129][3];
long long check(int n, int t) {
  int ans = 0;
  for (int i = 1; i <= n; ++i)
    if ((1 << (bit(i) - 1)) == t) ++ans;
  return ans;
}
long long solve(long long n, long long t) {
  if (t == 0) exit(-1);
  ++n;
  v.clear();
  while (n) {
    v.push_back(n % 2);
    n >>= 1;
  }
  reverse(v.begin(), v.end());
  dp[0][0][0] = 1;
  for (int i = 0; i < v.size(); ++i)
    for (int j = 0; j <= i + 1; ++j)
      for (int less = 0; less < 2; ++less) {
        for (int t = less ? 1 : v[i]; t >= 0; --t) {
          dp[i + 1][j + t][less ? 1 : (t < v[i])] += dp[i][j][less];
        }
      }
  int cnt = 1;
  while (t > 1) {
    if (t % 2 == 1) return 0;
    t >>= 1, cnt++;
  }
  return dp[v.size()][cnt][1];
}
int main() {
  long long n, t;
  cin >> n >> t;
  cout << solve(n + 1, t) - (t == 1) << 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;
int a[560][560] = {0};
int build(int m, bool out = true) {
  memset(a, 0, sizeof(a));
  for (int j = 1; j <= m; j++) a[1][j] = 0;
  a[1][m + 1] = 1;
  for (int i = 2; i <= m + 1; ++i)
    for (int j = 1; j <= m + 1; ++j)
      if (j == 1)
        a[i][j] = a[i - 1][j + 1];
      else {
        if (j == m + 1)
          a[i][j] = a[i - 1][j - 1];
        else
          a[i][j] = a[i - 1][j - 1] ^ a[i - 1][j + 1];
      }
  int ret = 0;
  if (out) cout << "For:	" << m << endl;
  for (int i = 1; i <= m + 1; ++i) {
    ret += a[m + 1][i];
    for (int j = 1; j <= m + 1; ++j)
      if (out) cout << a[i][j] << "	";
    if (out) cout << endl;
  }
  if (out) cout << endl;
  return ret;
}
int C(int n, int k) {
  int res = 0;
  for (int i = n - k + 1; i <= n; ++i) {
    int j = i;
    while (j % 2 == 0) ++res, j /= 2;
  }
  for (int i = 1; i <= k; ++i) {
    int j = i;
    while (j % 2 == 0) --res, j /= 2;
  }
  return !res;
}
int bit(int x) {
  int ret = 0;
  while (x) {
    ret += (x & 1);
    x >>= 1;
  }
  return ret;
}
vector<int> v;
long long dp[129][129][3];
long long check(int n, int t) {
  int ans = 0;
  for (int i = 1; i <= n; ++i)
    if ((1 << (bit(i) - 1)) == t) ++ans;
  return ans;
}
long long solve(long long n, long long t) {
  if (t == 0) exit(-1);
  ++n;
  v.clear();
  while (n) {
    v.push_back(n % 2);
    n >>= 1;
  }
  reverse(v.begin(), v.end());
  dp[0][0][0] = 1;
  for (int i = 0; i < v.size(); ++i)
    for (int j = 0; j <= i + 1; ++j)
      for (int less = 0; less < 2; ++less) {
        for (int t = less ? 1 : v[i]; t >= 0; --t) {
          dp[i + 1][j + t][less ? 1 : (t < v[i])] += dp[i][j][less];
        }
      }
  int cnt = 1;
  while (t > 1) {
    if (t % 2 == 1) return 0;
    t >>= 1, cnt++;
  }
  return dp[v.size()][cnt][1];
}
int main() {
  long long n, t;
  cin >> n >> t;
  cout << solve(n + 1, t) - (t == 1) << endl;
  return 0;
}

```