#include <bits/stdc++.h>
using namespace std;
const int N = 100000 + 7;
const int M = 48;
const int inf = 1e9 + 7;
const long long base = 1e18;
const double pi = acos(-1);
const double ep = 1e-9;
int a[50] = {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, 32582657};
long long power(int k) {
  if (k == 0) return 1;
  long long tmp = power(k / 2);
  tmp = (tmp * tmp) % inf;
  if (k & 1) tmp = tmp * 2 % inf;
  return tmp;
}
int main() {
  int n;
  while (cin >> n) {
    cout << power(a[n - 1] - 1) - 1 << endl;
  }
}

### Prompt

Please create a solution in cpp 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 N = 100000 + 7;
const int M = 48;
const int inf = 1e9 + 7;
const long long base = 1e18;
const double pi = acos(-1);
const double ep = 1e-9;
int a[50] = {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, 32582657};
long long power(int k) {
  if (k == 0) return 1;
  long long tmp = power(k / 2);
  tmp = (tmp * tmp) % inf;
  if (k & 1) tmp = tmp * 2 % inf;
  return tmp;
}
int main() {
  int n;
  while (cin >> n) {
    cout << power(a[n - 1] - 1) - 1 << endl;
  }
}

```