#include <bits/stdc++.h>
using namespace std;
long long n;
long long a[100] = {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};
long long Pow2(long long a, long long b) {
  long long ans = 1LL;
  for (long long i = 1; i <= b; i <<= 1, a = (a * a) % 1000000007)
    if (i & b) ans = (ans * a) % 1000000007;
  return ans;
}
int main() {
  scanf("%I64d", &n);
  printf("%I64d\n", (Pow2(2, a[n] - 1) - 1 + 1000000007) % 1000000007);
  return 0;
}

### Prompt

Develop a solution in cpp 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[100] = {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};
long long Pow2(long long a, long long b) {
  long long ans = 1LL;
  for (long long i = 1; i <= b; i <<= 1, a = (a * a) % 1000000007)
    if (i & b) ans = (ans * a) % 1000000007;
  return ans;
}
int main() {
  scanf("%I64d", &n);
  printf("%I64d\n", (Pow2(2, a[n] - 1) - 1 + 1000000007) % 1000000007);
  return 0;
}

```