#include <bits/stdc++.h>
using namespace std;
long long n;
long long a[45] = {0,       2,       3,       5,        7,       13,
                   17,      19,      31,      61,       89,      107,
                   127,     521,     607,     1279,     2203,    2281,
                   3217,    4253,    4423,    9689,     9941,    11213,
                   19937,   21701,   23209,   44497,    86243,   110503,
                   132049,  216091,  756839,  859433,   1257787, 1398269,
                   2976221, 3021377, 6972593, 13466917, 20996011};
long long ll(long long x, long long k) {
  long long b = 1;
  while (k > 0) {
    if (k & 1) b = (b * x) % 1000000007;
    x = (x * x) % 1000000007;
    k >>= 1;
  }
  return b;
}
int main() {
  while (~scanf("%lld", &n)) {
    printf("%lld\n", ll(2, a[n] - 1) - 1);
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
long long n;
long long a[45] = {0,       2,       3,       5,        7,       13,
                   17,      19,      31,      61,       89,      107,
                   127,     521,     607,     1279,     2203,    2281,
                   3217,    4253,    4423,    9689,     9941,    11213,
                   19937,   21701,   23209,   44497,    86243,   110503,
                   132049,  216091,  756839,  859433,   1257787, 1398269,
                   2976221, 3021377, 6972593, 13466917, 20996011};
long long ll(long long x, long long k) {
  long long b = 1;
  while (k > 0) {
    if (k & 1) b = (b * x) % 1000000007;
    x = (x * x) % 1000000007;
    k >>= 1;
  }
  return b;
}
int main() {
  while (~scanf("%lld", &n)) {
    printf("%lld\n", ll(2, a[n] - 1) - 1);
  }
  return 0;
}

```