#include <bits/stdc++.h>
using namespace std;
const int maxn = 70;
long long N, T, dn[maxn][maxn][2];
long long dfs(int node, int left, int less) {
  if (node == -1) return !left;
  long long &sum = dn[node][left][less];
  if (sum != -1) return sum;
  sum = 0;
  if ((less || N & (1ll << node)) && left) sum += dfs(node - 1, left - 1, less);
  sum += dfs(node - 1, left, !!(N & (1ll << node) | less));
  return sum;
}
int main() {
  scanf("%lld %lld", &N, &T);
  N++;
  int x = -1, last = 0;
  for (int i = 0; i <= 60; i++) {
    if (T == (1ll << i)) x = i;
    if (N & (1ll << i)) last = i;
  }
  if (x == -1) {
    printf("0\n");
    return 0;
  }
  memset(dn, -1, sizeof dn);
  printf("%lld\n", dfs(last, x + 1, 0) - (T == 1));
  return 0;
}

### Prompt

Develop a solution in cpp 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 maxn = 70;
long long N, T, dn[maxn][maxn][2];
long long dfs(int node, int left, int less) {
  if (node == -1) return !left;
  long long &sum = dn[node][left][less];
  if (sum != -1) return sum;
  sum = 0;
  if ((less || N & (1ll << node)) && left) sum += dfs(node - 1, left - 1, less);
  sum += dfs(node - 1, left, !!(N & (1ll << node) | less));
  return sum;
}
int main() {
  scanf("%lld %lld", &N, &T);
  N++;
  int x = -1, last = 0;
  for (int i = 0; i <= 60; i++) {
    if (T == (1ll << i)) x = i;
    if (N & (1ll << i)) last = i;
  }
  if (x == -1) {
    printf("0\n");
    return 0;
  }
  memset(dn, -1, sizeof dn);
  printf("%lld\n", dfs(last, x + 1, 0) - (T == 1));
  return 0;
}

```