#include <bits/stdc++.h>
const int mod = 1e9 + 7;
int n, pr[41] = {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};
int pow(int x) {
  long long ans = 1, base = 2;
  for (; x; x >>= 1) {
    if (x & 1) ans = ans * base % mod;
    base = base * base % mod;
  }
  return (int)ans;
}
int main() {
  scanf("%d", &n);
  printf("%d", pow(pr[n] - 1) - 1);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
const int mod = 1e9 + 7;
int n, pr[41] = {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};
int pow(int x) {
  long long ans = 1, base = 2;
  for (; x; x >>= 1) {
    if (x & 1) ans = ans * base % mod;
    base = base * base % mod;
  }
  return (int)ans;
}
int main() {
  scanf("%d", &n);
  printf("%d", pow(pr[n] - 1) - 1);
  return 0;
}

```