human
It is the easy version of the problem. The difference is that in this version, there are no nodes with already chosen colors.

Theofanis is starving, and he wants to eat his favorite food, sheftalia. However, he should first finish his homework. Can you help him with this problem?

You have a perfect binary tree of 2^k - 1 nodes — a binary tree where all vertices i from 1 to 2^{k - 1} - 1 have exactly two children: vertices 2i and 2i + 1. Vertices from 2^{k - 1} to 2^k - 1 don't have any children. You want to color its vertices with the 6 Rubik's cube colors (White, Green, Red, Blue, Orange and Yellow).

Let's call a coloring good when all edges connect nodes with colors that are neighboring sides in the Rubik's cube.

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A picture of Rubik's cube and its 2D map.

More formally: 

  * a white node can not be neighboring with white and yellow nodes.

  * a yellow node can not be neighboring with white and yellow nodes.

  * a green node can not be neighboring with green and blue nodes.

  * a blue node can not be neighboring with green and blue nodes.

  * a red node can not be neighboring with red and orange nodes.

  * an orange node can not be neighboring with red and orange nodes.



You want to calculate the number of the good colorings of the binary tree. Two colorings are considered different if at least one node is colored with a different color.

The answer may be too large, so output the answer modulo 10^9+7.

Input

The first and only line contains the integers k (1 ≤ k ≤ 60) — the number of levels in the perfect binary tree you need to color.

Output

Print one integer — the number of the different colorings modulo 10^9+7.

Examples

Input

3

Output

24576

Input

14

Output

934234

Note

In the picture below, you can see one of the correct colorings of the first example.

Using python3 can you solve the prior task?
 Give me a Python code to print one integer — the number of the different colorings modulo 10^9+7?
assistant
#Here is Python code to print one integer — the number of the different colorings modulo 10^9+7.

import os, sys
from io import BytesIO, IOBase
from types import GeneratorType
from bisect import *
from collections import defaultdict, deque, Counter
import math, string
from heapq import *
from operator import add
from itertools import accumulate

BUFSIZE = 8192
sys.setrecursionlimit(10 ** 5)

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        import os

        self.os = os
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            self.os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

inf = float("inf")
en = lambda x:...