#include <bits/stdc++.h>
using namespace std;
void hoge() {
  const int M = 110;
  vector<int> prv(M + 1), nxt(M + 1);
  prv[1] = 1;
  for (int i = 1; i < M; ++i) {
    for (int j = 1; j <= M; ++j) {
      if (j == 1) {
        nxt[j] = prv[j + 1];
      } else if (j == M) {
        nxt[j] = prv[j - 1];
      } else {
        nxt[j] = prv[j - 1] ^ prv[j + 1];
      }
    }
    int s = accumulate((nxt).begin(), (nxt).end(), 0);
    printf("%d: %d, %d\n", i, s, 1 << (__builtin_popcount(i + 1) - 1));
    prv.swap(nxt);
  }
}
long long cmb(long long n, long long k) {
  if (k < 0 || k > n) return 0;
  long long res = 1;
  for (int i = 0; i < k; ++i) res = res * (n - i) / (i + 1);
  return res;
}
long long solve(long long n, int p) {
  if (p == 0)
    return 1;
  else if (n == 0)
    return 0;
  else {
    int b = 63 - __builtin_clzll(n);
    long long l = 1LL << b;
    long long ans = solve(n - l, p - 1);
    ans += cmb(b, p);
    return ans;
  }
}
int main() {
  long long N, T;
  while (cin >> N >> T) {
    int P = 0;
    while (T > 1 && T % 2 == 0) {
      ++P;
      T /= 2;
    }
    if (T != 1) {
      puts("0");
      continue;
    }
    long long ans = solve(N + 1, P + 1) - solve(1, P + 1);
    cout << ans << 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>
using namespace std;
void hoge() {
  const int M = 110;
  vector<int> prv(M + 1), nxt(M + 1);
  prv[1] = 1;
  for (int i = 1; i < M; ++i) {
    for (int j = 1; j <= M; ++j) {
      if (j == 1) {
        nxt[j] = prv[j + 1];
      } else if (j == M) {
        nxt[j] = prv[j - 1];
      } else {
        nxt[j] = prv[j - 1] ^ prv[j + 1];
      }
    }
    int s = accumulate((nxt).begin(), (nxt).end(), 0);
    printf("%d: %d, %d\n", i, s, 1 << (__builtin_popcount(i + 1) - 1));
    prv.swap(nxt);
  }
}
long long cmb(long long n, long long k) {
  if (k < 0 || k > n) return 0;
  long long res = 1;
  for (int i = 0; i < k; ++i) res = res * (n - i) / (i + 1);
  return res;
}
long long solve(long long n, int p) {
  if (p == 0)
    return 1;
  else if (n == 0)
    return 0;
  else {
    int b = 63 - __builtin_clzll(n);
    long long l = 1LL << b;
    long long ans = solve(n - l, p - 1);
    ans += cmb(b, p);
    return ans;
  }
}
int main() {
  long long N, T;
  while (cin >> N >> T) {
    int P = 0;
    while (T > 1 && T % 2 == 0) {
      ++P;
      T /= 2;
    }
    if (T != 1) {
      puts("0");
      continue;
    }
    long long ans = solve(N + 1, P + 1) - solve(1, P + 1);
    cout << ans << endl;
  }
  return 0;
}

```