#include <bits/stdc++.h>
using namespace std;
const int maxn = 60;
const long long mod = 1e9 + 7;
int a[] = {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, 37156667, 42643801, 43112609};
long long fpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
int main() {
  int z;
  cin >> z;
  cout << fpow(2, a[z - 1] - 1) - 1 << "\n";
  return 0;
}

### 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 maxn = 60;
const long long mod = 1e9 + 7;
int a[] = {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, 37156667, 42643801, 43112609};
long long fpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
int main() {
  int z;
  cin >> z;
  cout << fpow(2, a[z - 1] - 1) - 1 << "\n";
  return 0;
}

```