#include <bits/stdc++.h>
const int N = 1010;
using namespace std;
void read(long long &x) {
  x = 0;
  char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    ;
  for (; c >= '0' && c <= '9'; c = getchar())
    x = (x << 3) + (x << 1) + (c ^ 48);
}
long long n, t;
long long cnt[N][N], g[N];
int len = 0;
long long CHK() {
  len = 0;
  while (t != 1) {
    if (t & 1) return -1;
    len++;
    t /= 2;
  }
  return 0;
}
int LG(long long now) { return (int)log2(now); }
long long GET(long long lef, int hlp) {
  if (len < hlp || lef == 0) return 0;
  if (lef == 1) return (hlp == len);
  int tmp = LG(lef);
  long long now = g[tmp];
  if (now == lef) return cnt[tmp][len - hlp];
  return cnt[tmp][len - hlp] + GET(lef - now, hlp + 1);
}
void INIT() {
  g[0] = 1;
  for (int i = 1; i <= 50; i++) g[i] = g[i - 1] * 2;
}
long long SOLVE() {
  if (CHK() == -1) return 0;
  long long ans = 0;
  long long lef = n;
  cnt[0][0] = 1;
  for (int i = 1; i <= 50; i++) {
    long long now = g[i];
    if (now > lef) return ans + GET(lef, 0);
    lef -= now;
    for (int j = 0; j <= len; j++) {
      cnt[i][j] = cnt[i - 1][j];
      if (j > 0) {
        cnt[i][j] += cnt[i - 1][j - 1];
      }
    }
    ans += cnt[i][len];
  }
  return ans;
}
int main() {
  read(n);
  read(t);
  INIT();
  cout << SOLVE() << endl;
  return 0;
}

### Prompt

Please formulate 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>
const int N = 1010;
using namespace std;
void read(long long &x) {
  x = 0;
  char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    ;
  for (; c >= '0' && c <= '9'; c = getchar())
    x = (x << 3) + (x << 1) + (c ^ 48);
}
long long n, t;
long long cnt[N][N], g[N];
int len = 0;
long long CHK() {
  len = 0;
  while (t != 1) {
    if (t & 1) return -1;
    len++;
    t /= 2;
  }
  return 0;
}
int LG(long long now) { return (int)log2(now); }
long long GET(long long lef, int hlp) {
  if (len < hlp || lef == 0) return 0;
  if (lef == 1) return (hlp == len);
  int tmp = LG(lef);
  long long now = g[tmp];
  if (now == lef) return cnt[tmp][len - hlp];
  return cnt[tmp][len - hlp] + GET(lef - now, hlp + 1);
}
void INIT() {
  g[0] = 1;
  for (int i = 1; i <= 50; i++) g[i] = g[i - 1] * 2;
}
long long SOLVE() {
  if (CHK() == -1) return 0;
  long long ans = 0;
  long long lef = n;
  cnt[0][0] = 1;
  for (int i = 1; i <= 50; i++) {
    long long now = g[i];
    if (now > lef) return ans + GET(lef, 0);
    lef -= now;
    for (int j = 0; j <= len; j++) {
      cnt[i][j] = cnt[i - 1][j];
      if (j > 0) {
        cnt[i][j] += cnt[i - 1][j - 1];
      }
    }
    ans += cnt[i][len];
  }
  return ans;
}
int main() {
  read(n);
  read(t);
  INIT();
  cout << SOLVE() << endl;
  return 0;
}

```