Snuke lives in another world, where slimes are real creatures and kept by some people. Slimes come in N colors. Those colors are conveniently numbered 1 through N. Snuke currently has no slime. His objective is to have slimes of all the colors together.

Snuke can perform the following two actions:

* Select a color i (1≤i≤N), such that he does not currently have a slime in color i, and catch a slime in color i. This action takes him a_i seconds.

* Cast a spell, which changes the color of all the slimes that he currently has. The color of a slime in color i (1≤i≤N-1) will become color i+1, and the color of a slime in color N will become color 1. This action takes him x seconds.




Find the minimum time that Snuke needs to have slimes in all N colors.

Constraints

* 2≤N≤2,000
* a_i are integers.
* 1≤a_i≤10^9
* x is an integer.
* 1≤x≤10^9

Input

The input is given from Standard Input in the following format:


N x
a_1 a_2 ... a_N


Output

Find the minimum time that Snuke needs to have slimes in all N colors.

Examples

Input

2 10
1 100


Output

12


Input

3 10
100 1 100


Output

23


Input

4 10
1 2 3 4


Output

10
# -*- coding: utf-8 -*-
import math,string,itertools,fractions,heapq,collections,re,array,bisect

def s():
    return raw_input().strip()
def n():
    return int(raw_input())
def d():
    return float(raw_input())

def ls():
    return raw_input().strip().split()
def ln():
    return map(int, raw_input().strip().split())
def ld():
    return map(float, raw_input().strip().split())

def fs():
    return [raw_input().strip() for i in xrange(input())]
def fn():
    return [int(raw_input().strip()) for i in xrange(input())]
def fd():
    return [float(raw_input().strip()) for i in xrange(input())]

EPS = 1e-9

#http://www.deqnotes.net/acmicpc/2d_geometry/lines
def dot(a, b):
    return a.real * b.real + a.imag * b.imag

def cross(a, b):
    return a.real * b.imag - a.imag * b.real

def distance(a, b, c):
    if dot(b-a, c-a) <= EPS:
        return abs(c-a)
    if dot(a-b, c-b) <= EPS:
        return abs(c-b)
    return abs(cross(b-a, c-a)/(b-a))

N, x = ln()

A = ln()

result = 0
m_d_all = 0
for j in xrange(N):
    m = 1e10
    m_i = -1
    m_d = -1
    m_raw = 0
    for i in xrange(N):
        distance = j - i if j >= i else N - i + j
        cost = distance * x + A[i]
        if cost < m:
            m = cost
            m_i = j
            m_raw = A[i]
            m_d = distance
        if cost == m:
            m_d = min(distance, m_d)
    if m_d > m_d_all:
        m_d_all = m_d

    result += m_raw

print result + m_d_all * x
