Two Formats

A byte is a unit of information made of eight binary digits. One byte can therefore hold the $2^8$ integers from $0$ to $255$. To store larger numbers, a computer uses several bytes. The number is written in base $256$, and its digits are stored in the bytes. That is, to store $A = \sum_{i=0}^{k} 256^i \cdot A_i$ ($0 \le A_i \le 255$), $(k+1)$ bytes holding $A_0, A_1, \ldots, A_k$ are used. For example, storing $2000000$ takes 3 bytes: $2000000 = 128 \cdot 256^0 + 132 \cdot 256^1 + 30 \cdot 256^2$, so the bytes hold $128$, $132$, and $30$.

Devices differ in the order in which they store the bytes of one number. The two most common orders are big-endian and little-endian. Big-endian stores the bytes from the most significant digit to the least, in the order $A_k, A_{k-1}, \ldots, A_1, A_0$. For $2000000$, this order is $30, 132, 128$. Little-endian is the reverse of big-endian.

Now suppose a number is sent from a big-endian device to a little-endian device. The receiving device reads the bytes in its own order, so it may read a completely different number. For example, sending $2000000$ transmits the bytes $30, 132, 128$. The receiver reads them as $30 \cdot 256^0 + 132 \cdot 256^1 + 128 \cdot 256^2 = 8422430$.

You are given the number $N$ recorded on the little-endian device after the number $M$ was received from the big-endian device. Recover $M$.

The first line contains one integer $N$ ($1 \le N < 2^{24}$).

Print the number $M$.