#include <bits/stdc++.h>
using namespace std;
const long long module = 1000000007LL;
long long dat[40] = {
    1,      2,       4,       6,       12,      16,      18,       30,
    60,     88,      106,     126,     520,     606,     1278,     2202,
    2280,   3216,    4252,    4422,    9688,    9940,    11212,    19936,
    21700,  23208,   44496,   86242,   110502,  132048,  216090,   756838,
    859432, 1257786, 1398268, 2976220, 3021376, 6972592, 13466916, 20996010};
long long p2(long long x) {
  long long s = 1, t = 2;
  while (x) {
    if (x & 1) s = (s * t) % module;
    t = (t * t) % module;
    x >>= 1;
  }
  return s;
}
int main() {
  long long x;
  cin >> x;
  cout << (p2(dat[x - 1]) - 1 + module) % module;
  return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

Consider the following equation: 

<image> where sign [a] represents the integer part of number a.

Let's find all integer z (z > 0), for which this equation is unsolvable in positive integers. The phrase "unsolvable in positive integers" means that there are no such positive integers x and y (x, y > 0), for which the given above equation holds.

Let's write out all such z in the increasing order: z1, z2, z3, and so on (zi < zi + 1). Your task is: given the number n, find the number zn.

Input

The first line contains a single integer n (1 ≤ n ≤ 40).

Output

Print a single integer — the number zn modulo 1000000007 (109 + 7). It is guaranteed that the answer exists.

Examples

Input

1


Output

1

Input

2


Output

3

Input

3


Output

15

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long module = 1000000007LL;
long long dat[40] = {
    1,      2,       4,       6,       12,      16,      18,       30,
    60,     88,      106,     126,     520,     606,     1278,     2202,
    2280,   3216,    4252,    4422,    9688,    9940,    11212,    19936,
    21700,  23208,   44496,   86242,   110502,  132048,  216090,   756838,
    859432, 1257786, 1398268, 2976220, 3021376, 6972592, 13466916, 20996010};
long long p2(long long x) {
  long long s = 1, t = 2;
  while (x) {
    if (x & 1) s = (s * t) % module;
    t = (t * t) % module;
    x >>= 1;
  }
  return s;
}
int main() {
  long long x;
  cin >> x;
  cout << (p2(dat[x - 1]) - 1 + module) % module;
  return 0;
}

```