#include <bits/stdc++.h>
using namespace std;
vector<int> v;
long long dp[50][50][2];
int ok(long long x) {
  int ret = 0;
  for (; x > 1;) {
    if (x & 1) return 0;
    x >>= 1;
    ret++;
  }
  return ret + 1;
}
long long dfs(int id, int c, int tag, int need) {
  if (dp[id][c][tag] >= 0) return dp[id][c][tag];
  long long ret = 0;
  if (id == v.size()) {
    if (c == need)
      ret = 1;
    else
      ret = 0;
    return dp[id][c][tag] = ret;
  }
  if (tag) {
    ret += dfs(id + 1, c, tag, need);
    if (c < need) ret += dfs(id + 1, c + 1, tag, need);
  } else {
    if (v[id] == 1) {
      ret += dfs(id + 1, c, 1, need);
      if (c < need) ret += dfs(id + 1, c + 1, tag, need);
    } else
      ret += dfs(id + 1, c, tag, need);
  }
  return dp[id][c][tag] = ret;
}
void gao(int cnt, long long n) {
  v.clear();
  for (; n; n >>= 1) v.push_back(n & 1);
  reverse(v.begin(), v.end());
  memset(dp, -1, sizeof(dp));
  cout << dfs(0, 0, 0, cnt) - (cnt == 1) << endl;
}
int main() {
  long long n, t;
  while (cin >> n >> t) {
    int cnt = ok(t);
    if (cnt) {
      gao(cnt, n + 1);
    } else
      puts("0");
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
vector<int> v;
long long dp[50][50][2];
int ok(long long x) {
  int ret = 0;
  for (; x > 1;) {
    if (x & 1) return 0;
    x >>= 1;
    ret++;
  }
  return ret + 1;
}
long long dfs(int id, int c, int tag, int need) {
  if (dp[id][c][tag] >= 0) return dp[id][c][tag];
  long long ret = 0;
  if (id == v.size()) {
    if (c == need)
      ret = 1;
    else
      ret = 0;
    return dp[id][c][tag] = ret;
  }
  if (tag) {
    ret += dfs(id + 1, c, tag, need);
    if (c < need) ret += dfs(id + 1, c + 1, tag, need);
  } else {
    if (v[id] == 1) {
      ret += dfs(id + 1, c, 1, need);
      if (c < need) ret += dfs(id + 1, c + 1, tag, need);
    } else
      ret += dfs(id + 1, c, tag, need);
  }
  return dp[id][c][tag] = ret;
}
void gao(int cnt, long long n) {
  v.clear();
  for (; n; n >>= 1) v.push_back(n & 1);
  reverse(v.begin(), v.end());
  memset(dp, -1, sizeof(dp));
  cout << dfs(0, 0, 0, cnt) - (cnt == 1) << endl;
}
int main() {
  long long n, t;
  while (cin >> n >> t) {
    int cnt = ok(t);
    if (cnt) {
      gao(cnt, n + 1);
    } else
      puts("0");
  }
  return 0;
}

```