#include <bits/stdc++.h>
#pragma GCC target("sse4")
#pragma GCC target("avx,tune=native")
using namespace std;
long long n, t;
long long dp[51][51];
long long twopow[51];
long long count(long long number, long long bits) {
  if (bits == 0) return 1;
  if (number == 1) return (bits == 1 ? 1 : 0);
  if (number == 0) return (bits >= 1 ? 0 : 1);
  int firstBit = 0;
  for (int i = 49; i >= 0; i--) {
    if (twopow[i] <= number) {
      firstBit = i;
      break;
    }
  }
  if (bits > firstBit + 1) return 0;
  long long answer = dp[firstBit][bits];
  answer += count(number - twopow[firstBit], bits - 1);
  return answer;
}
int main() {
  if (fopen("FILENAME.in", "r")) {
    freopen("FILENAME.in", "r", stdin);
    freopen("FILENAME.out", "w", stdout);
  }
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> t;
  dp[0][0] = 1;
  for (int i = 0; i < 50; i++) {
    for (int j = 0; j < 50; j++) {
      dp[i + 1][j + 1] += dp[i][j];
      dp[i + 1][j] += dp[i][j];
    }
  }
  twopow[0] = 1LL;
  int bitsNeeded = -1;
  for (int i = 0; i < 50; i++) {
    if (twopow[i] == t) {
      bitsNeeded = i + 1;
    }
    twopow[i + 1] = twopow[i] * 2;
  }
  if (bitsNeeded == -1) {
    cout << 0 << endl;
    return 0;
  }
  cout << count(n + 1, bitsNeeded) - count(1, bitsNeeded) << endl;
}

### 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>
#pragma GCC target("sse4")
#pragma GCC target("avx,tune=native")
using namespace std;
long long n, t;
long long dp[51][51];
long long twopow[51];
long long count(long long number, long long bits) {
  if (bits == 0) return 1;
  if (number == 1) return (bits == 1 ? 1 : 0);
  if (number == 0) return (bits >= 1 ? 0 : 1);
  int firstBit = 0;
  for (int i = 49; i >= 0; i--) {
    if (twopow[i] <= number) {
      firstBit = i;
      break;
    }
  }
  if (bits > firstBit + 1) return 0;
  long long answer = dp[firstBit][bits];
  answer += count(number - twopow[firstBit], bits - 1);
  return answer;
}
int main() {
  if (fopen("FILENAME.in", "r")) {
    freopen("FILENAME.in", "r", stdin);
    freopen("FILENAME.out", "w", stdout);
  }
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> t;
  dp[0][0] = 1;
  for (int i = 0; i < 50; i++) {
    for (int j = 0; j < 50; j++) {
      dp[i + 1][j + 1] += dp[i][j];
      dp[i + 1][j] += dp[i][j];
    }
  }
  twopow[0] = 1LL;
  int bitsNeeded = -1;
  for (int i = 0; i < 50; i++) {
    if (twopow[i] == t) {
      bitsNeeded = i + 1;
    }
    twopow[i + 1] = twopow[i] * 2;
  }
  if (bitsNeeded == -1) {
    cout << 0 << endl;
    return 0;
  }
  cout << count(n + 1, bitsNeeded) - count(1, bitsNeeded) << endl;
}

```