#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
int a[44] = {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, 24036583,
             25964951, 30402457};
long long pow(long long n, int k) {
  long long ans = 1;
  while (k) {
    if (k % 2) ans = (ans * n) % mod;
    n = (n * n) % mod;
    k /= 2;
  }
  return ans % mod;
}
int main() {
  int n;
  scanf("%d", &n);
  printf("%d\n", (pow(2ll, a[n] - 1) - 1 + mod) % mod);
}

### Prompt

Your challenge is to write a Cpp solution to 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 int mod = 1000000007;
int a[44] = {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, 24036583,
             25964951, 30402457};
long long pow(long long n, int k) {
  long long ans = 1;
  while (k) {
    if (k % 2) ans = (ans * n) % mod;
    n = (n * n) % mod;
    k /= 2;
  }
  return ans % mod;
}
int main() {
  int n;
  scanf("%d", &n);
  printf("%d\n", (pow(2ll, a[n] - 1) - 1 + mod) % mod);
}

```