#include <bits/stdc++.h>
using namespace std;
const int mod = 10000007;
const int maxn = 55;
long long n;
int cnt = 0;
int num[55];
long long dp[maxn][maxn][2];
long long pow_mod(long long x, long long p) {
  long long res = 1;
  while (p) {
    if (p & 1) res = (res * x) % mod;
    x = (x * x) % mod;
    p >>= 1;
  }
  return res;
}
int main() {
  long long t;
  int tt = 0;
  scanf("%lld", &n);
  scanf("%lld", &t);
  if (t != (t & -t)) {
    printf("0\n");
    return 0;
  } else {
    while (t) {
      ++tt;
      t /= 2;
    }
  }
  n++;
  while (n) {
    ++cnt;
    if (n & 1) num[cnt] = 1;
    n >>= 1;
  }
  dp[cnt][1][1] = 1;
  dp[cnt][0][0] = 1;
  for (register int i = cnt - 1; i >= 1; i--) {
    if (num[i] == 1) {
      for (register int j = 0; j <= cnt - i + 1; j++) {
        dp[i][j][1] += dp[i + 1][j - 1][1];
        dp[i][j][0] += dp[i + 1][j][1];
        dp[i][j][0] += dp[i + 1][j][0];
        dp[i][j][0] += dp[i + 1][j - 1][0];
      }
    } else {
      for (register int j = 0; j <= cnt - i + 1; j++) {
        dp[i][j][1] += dp[i + 1][j][1];
        dp[i][j][0] += dp[i + 1][j][0];
        dp[i][j][0] += dp[i + 1][j - 1][0];
      }
    }
  }
  dp[1][1][1]--;
  printf("%lld\n", dp[1][tt][1] + dp[1][tt][0]);
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 mod = 10000007;
const int maxn = 55;
long long n;
int cnt = 0;
int num[55];
long long dp[maxn][maxn][2];
long long pow_mod(long long x, long long p) {
  long long res = 1;
  while (p) {
    if (p & 1) res = (res * x) % mod;
    x = (x * x) % mod;
    p >>= 1;
  }
  return res;
}
int main() {
  long long t;
  int tt = 0;
  scanf("%lld", &n);
  scanf("%lld", &t);
  if (t != (t & -t)) {
    printf("0\n");
    return 0;
  } else {
    while (t) {
      ++tt;
      t /= 2;
    }
  }
  n++;
  while (n) {
    ++cnt;
    if (n & 1) num[cnt] = 1;
    n >>= 1;
  }
  dp[cnt][1][1] = 1;
  dp[cnt][0][0] = 1;
  for (register int i = cnt - 1; i >= 1; i--) {
    if (num[i] == 1) {
      for (register int j = 0; j <= cnt - i + 1; j++) {
        dp[i][j][1] += dp[i + 1][j - 1][1];
        dp[i][j][0] += dp[i + 1][j][1];
        dp[i][j][0] += dp[i + 1][j][0];
        dp[i][j][0] += dp[i + 1][j - 1][0];
      }
    } else {
      for (register int j = 0; j <= cnt - i + 1; j++) {
        dp[i][j][1] += dp[i + 1][j][1];
        dp[i][j][0] += dp[i + 1][j][0];
        dp[i][j][0] += dp[i + 1][j - 1][0];
      }
    }
  }
  dp[1][1][1]--;
  printf("%lld\n", dp[1][tt][1] + dp[1][tt][0]);
  return 0;
}

```