#include <bits/stdc++.h>
using namespace std;
int ToInt(string& s, int len) {
  int r = 0;
  for (int i = 0, max_ = (len); i < max_; ++i) {
    r *= 10;
    r += s[i] - '0';
  }
  return r;
}
int GCD(int a, int b) { return b != 0 ? GCD(b, a % b) : a; }
int LCM(int a, int b) { return a * (b / GCD(a, b)); }
long long Pow(long long n, long long e) {
  if (e == 0) return 1;
  if (e == 1)
    return n;
  else if (e & 1) {
    long long t = Pow(n, e / 2);
    return n * t * t;
  } else {
    long long t = Pow(n, e / 2);
    return t * t;
  }
}
const int MAX = 50;
long long comb[MAX][MAX];
void CombAll() {
  comb[0][0] = 1;
  comb[1][0] = 1;
  comb[1][1] = 1;
  for (int i = (2), max_ = (MAX); i < max_; ++i) {
    comb[i][0] = 1;
    for (int j = (1), max_ = (i); j < max_; ++j)
      comb[i][j] = comb[i - 1][j] + comb[i - 1][j - 1];
    comb[i][i] = 1;
  }
}
vector<int> v;
long long DP(int id, int rem) {
  if (rem > (int)(v).size() - id) return 0;
  if (id == (int)(v).size() - 1) {
    if (v[id] == 0) {
      if (rem > 0) return 0;
      return 1;
    } else {
      if (rem > 1) return 0;
      return 1;
    }
  }
  if (v[id] == 1) {
    long long res = 0;
    if (rem > 0) res += DP(id + 1, rem - 1);
    res += comb[(int)(v).size() - id - 1][rem];
    return res;
  } else {
    return DP(id + 1, rem);
  }
}
int main() {
  int B = 0;
  long long N, T, M;
  scanf("%I64d %I64d", &N, &T);
  if (T & (T - 1)) {
    printf("0\n");
    return 0;
  }
  ++N;
  while (N > 0) {
    v.push_back(N % 2);
    N /= 2;
  }
  reverse((v).begin(), (v).end());
  for (int i = 0, max_ = (50); i < max_; ++i)
    if (T == (1LL << i)) {
      B = i + 1;
      break;
    }
  CombAll();
  long long res = DP(0, B);
  if (B == 1) --res;
  printf("%I64d\n", res);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
int ToInt(string& s, int len) {
  int r = 0;
  for (int i = 0, max_ = (len); i < max_; ++i) {
    r *= 10;
    r += s[i] - '0';
  }
  return r;
}
int GCD(int a, int b) { return b != 0 ? GCD(b, a % b) : a; }
int LCM(int a, int b) { return a * (b / GCD(a, b)); }
long long Pow(long long n, long long e) {
  if (e == 0) return 1;
  if (e == 1)
    return n;
  else if (e & 1) {
    long long t = Pow(n, e / 2);
    return n * t * t;
  } else {
    long long t = Pow(n, e / 2);
    return t * t;
  }
}
const int MAX = 50;
long long comb[MAX][MAX];
void CombAll() {
  comb[0][0] = 1;
  comb[1][0] = 1;
  comb[1][1] = 1;
  for (int i = (2), max_ = (MAX); i < max_; ++i) {
    comb[i][0] = 1;
    for (int j = (1), max_ = (i); j < max_; ++j)
      comb[i][j] = comb[i - 1][j] + comb[i - 1][j - 1];
    comb[i][i] = 1;
  }
}
vector<int> v;
long long DP(int id, int rem) {
  if (rem > (int)(v).size() - id) return 0;
  if (id == (int)(v).size() - 1) {
    if (v[id] == 0) {
      if (rem > 0) return 0;
      return 1;
    } else {
      if (rem > 1) return 0;
      return 1;
    }
  }
  if (v[id] == 1) {
    long long res = 0;
    if (rem > 0) res += DP(id + 1, rem - 1);
    res += comb[(int)(v).size() - id - 1][rem];
    return res;
  } else {
    return DP(id + 1, rem);
  }
}
int main() {
  int B = 0;
  long long N, T, M;
  scanf("%I64d %I64d", &N, &T);
  if (T & (T - 1)) {
    printf("0\n");
    return 0;
  }
  ++N;
  while (N > 0) {
    v.push_back(N % 2);
    N /= 2;
  }
  reverse((v).begin(), (v).end());
  for (int i = 0, max_ = (50); i < max_; ++i)
    if (T == (1LL << i)) {
      B = i + 1;
      break;
    }
  CombAll();
  long long res = DP(0, B);
  if (B == 1) --res;
  printf("%I64d\n", res);
  return 0;
}

```