#include <bits/stdc++.h>
long long a[40] = {
    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};
long long pow(long long a, long long b, long long mod) {
  long long r = 1;
  while (b) {
    if (b & 1) r = r * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return r;
}
int main() {
  int n;
  long long mod = 1000000007;
  scanf("%d", &n);
  printf("%I64d", (pow(2, a[n - 1] - 1, mod) + mod - 1) % mod);
  return 0;
}

### Prompt

In cpp, your task is to solve 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>
long long a[40] = {
    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};
long long pow(long long a, long long b, long long mod) {
  long long r = 1;
  while (b) {
    if (b & 1) r = r * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return r;
}
int main() {
  int n;
  long long mod = 1000000007;
  scanf("%d", &n);
  printf("%I64d", (pow(2, a[n - 1] - 1, mod) + mod - 1) % mod);
  return 0;
}

```