Problem: One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.

Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.

Input

The first line contains the single integer n (1 ≤ n ≤ 100). The second line contains n space-separated integers ri (1 ≤ ri ≤ 1000) — the circles' radii. It is guaranteed that all circles are different.

Output

Print the single real number — total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10 - 4.

Examples

Input

1
1


Output

3.1415926536


Input

3
1 4 2


Output

40.8407044967

Note

In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals π × 12 = π.

In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (π × 42 - π × 22) + π × 12 = π × 12 + π = 13π

Solution (in PYTHON3):

Solution:
import sys
import os
from bisect import bisect_right as br  # c++ upperbound
from bisect import bisect_left as bl  # c++ lowerbound bl(array,element)
from io import IOBase, BytesIO
import heapq
import math
import collections
import itertools
import bisect
from sys import stdin, stdout
mod = 10 ** 9 + 7
pie = 3.1415926536
#import resource
#resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])
#import threading
# threading.stack_size(2**27)
#import sys
# sys.setrecursionlimit(10**6)
# fact=[1]
# for i in range(1,1000001):
#    fact.append((fact[-1]*i)%mod)
# ifact=[0]*1000001
# ifact[1000000]=pow(fact[1000000],mod-2,mod)
# for i in range(1000000,0,-1):
#    ifact[i-1]=(i*ifact[i])%mod
#from random import randint as rn
#from Queue import Queue as Q


def modinv(n, p):
    return pow(n, p-2, p)


def ncr(n, r, p):  # for using this uncomment the lines calculating fact and ifact
    t = ((fact[n])*((ifact[r]*ifact[n-r]) % p)) % p
    return t


def ain():  # takes array as input
    return list(map(int, sin().split()))


def sin():
    return input().strip()


def GCD(x, y):
    while(y):
        x, y = y, x % y
    return x


def readIntArray(size):
    # return list(map(int, input().strip().split()))[:size]
    arr = []
    for i in range(0, size):
        temp = int(sin().split())
        arr.append(temp)
    return arr


def readStringArray(size):
    return list(map(str, input().strip().split()))[:size]


def readFloatArray(size):
    return list(map(float, input().strip().split()))[:size]


def read2DIntArray(row, col):
    return [[int(input()) for x in range(col)] for y in range(row)]


def read2DStringArray(row, col):
    return [[str(input()) for x in range(col)] for y in range(row)]


def read2DFloatArray(row, col):
    return [[float(input()) for x in range(col)] for y in range(row)]


"""*******************************************************"""


def main():
    n = int(sin())
    arr = ain()
    arr = sorted(arr, reverse=True)
    area = 0

    for i in range(0, n, 2):
        if (i == n - 1):
            area += pie * (arr[i] ** 2)
        else:
            area += (pie * (arr[i] ** 2)) - (pie * (arr[i + 1] ** 2))

    print(area)


# Python 2 and 3 footer by Pajenegod and c1729
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange


BUFSIZE = 8192


class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0, 2),
                   super(FastIO, self).write(s))[0])
        return s

    def read(self):
        while self._fill():
            pass
        return super(FastIO, self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill()
            self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

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


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            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)


def input(): return sys.stdin.readline().rstrip('\r\n')


if __name__ == '__main__':
    main()
# threading.Thread(target=main).start()