#include <bits/stdc++.h>
using namespace std;
double EPS = 1e-9;
int MAXINT = 2147483647;
int MININT = -2147483647;
long long mod = 1000000007;
int pangkat[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 bigmod(long long k) {
  if (k == 1) return 2;
  if (k == 0) return 1;
  long long res = bigmod(k / 2);
  if (k % 2 == 0) {
    return (res * res) % mod;
  } else {
    return ((2 * res) % mod * res) % mod;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  long long yea = bigmod(pangkat[n - 1]) - 1;
  if (yea < 0) yea += mod;
  printf("%I64d\n", yea);
}

### 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;
double EPS = 1e-9;
int MAXINT = 2147483647;
int MININT = -2147483647;
long long mod = 1000000007;
int pangkat[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 bigmod(long long k) {
  if (k == 1) return 2;
  if (k == 0) return 1;
  long long res = bigmod(k / 2);
  if (k % 2 == 0) {
    return (res * res) % mod;
  } else {
    return ((2 * res) % mod * res) % mod;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  long long yea = bigmod(pangkat[n - 1]) - 1;
  if (yea < 0) yea += mod;
  printf("%I64d\n", yea);
}

```