#include <bits/stdc++.h>
using namespace std;
const int N = 100 + 7;
const int M = 19;
const int mid = M / 2;
const int mod = 1e9 + 9;
const int inf = 1e9 + 7;
const long long linf = 1ll * inf * inf;
const double pi = acos(-1);
const double eps = 1e-7;
const double ep = 1e-5;
const int maxn = 1e5 + 7;
const double PI = acos(-1);
int a[N][N];
int m;
map<pair<int, pair<long long, long long> >, long long> ans;
long long get(int type, long long n, long long t) {
  pair<int, pair<long long, long long> > tmp = make_pair(type, make_pair(n, t));
  if (ans.count(tmp)) return ans[tmp];
  if (type == 0) {
    return (t == 1);
  }
  long long lt = 1ll << (type - 1);
  if (n < lt) {
    return ans[tmp] = get(type - 1, n, t);
  } else {
    if (t & 1) return ans[tmp] = get(type - 1, lt - 1, t);
    return ans[tmp] = get(type - 1, n - lt, t / 2) + get(type - 1, lt - 1, t);
  }
}
void solve() {
  long long n, t;
  cin >> n >> t;
  long long lt = 1;
  long long res = 0;
  int type = 0;
  ++n;
  while (n >= lt) {
    res += get(type, lt - 1, t);
    n -= lt;
    lt *= 2;
    type++;
  }
  if (n > 0) {
    res += get(type, n - 1, t);
  }
  if (t == 1) {
    res--;
  }
  cout << res;
}
int main() {
  int T = 1;
  for (int i = (1); i < (T + 1); i++) {
    solve();
  }
}

### Prompt

Your challenge is to write a Cpp solution 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;
const int N = 100 + 7;
const int M = 19;
const int mid = M / 2;
const int mod = 1e9 + 9;
const int inf = 1e9 + 7;
const long long linf = 1ll * inf * inf;
const double pi = acos(-1);
const double eps = 1e-7;
const double ep = 1e-5;
const int maxn = 1e5 + 7;
const double PI = acos(-1);
int a[N][N];
int m;
map<pair<int, pair<long long, long long> >, long long> ans;
long long get(int type, long long n, long long t) {
  pair<int, pair<long long, long long> > tmp = make_pair(type, make_pair(n, t));
  if (ans.count(tmp)) return ans[tmp];
  if (type == 0) {
    return (t == 1);
  }
  long long lt = 1ll << (type - 1);
  if (n < lt) {
    return ans[tmp] = get(type - 1, n, t);
  } else {
    if (t & 1) return ans[tmp] = get(type - 1, lt - 1, t);
    return ans[tmp] = get(type - 1, n - lt, t / 2) + get(type - 1, lt - 1, t);
  }
}
void solve() {
  long long n, t;
  cin >> n >> t;
  long long lt = 1;
  long long res = 0;
  int type = 0;
  ++n;
  while (n >= lt) {
    res += get(type, lt - 1, t);
    n -= lt;
    lt *= 2;
    type++;
  }
  if (n > 0) {
    res += get(type, n - 1, t);
  }
  if (t == 1) {
    res--;
  }
  cout << res;
}
int main() {
  int T = 1;
  for (int i = (1); i < (T + 1); i++) {
    solve();
  }
}

```