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
import heapq

N, x = map(int, input().split())

INF = 10 ** 15
G = [[] for i in range(N+1)]
check, dist = [0 for j in range(N+1)], [INF for k in range(N+1)]
p, child = [-1 for l in range(N+1)], [-1 for m in range(N+1)]


def dijkstra(s):
    dist[s], pq = 0, []
    heapq.heappush(pq, (0, s))

    while len(pq) != 0:
        u = heapq.heappop(pq)
        check[u[1]] = 2

        if dist[u[1]] < u[0]:
            continue

        for v in G[u[1]]:
            if check[v[0]] != 2:
                if dist[u[1]] + v[1] < dist[v[0]]:
                    dist[v[0]], check[v[0]] = dist[u[1]] + v[1], 1
                    p[v[0]] = u[1]
                    heapq.heappush(pq, (dist[v[0]], v[0]))

    for i in range(N+1):
        if p[i] != -1:
            child[p[i]] = i


def main():
    a = list(map(int, input().split()))

    for i in range(N):
        G[0].append((i+1, a[i]))
        if i < N - 1:
            G[i+1].append((i+2, x))
    G[N].append((1, x))

    dijkstra(0)

    ans = 0
    for i in range(1, N+1):
        if p[i] == 0:
            v, cnt = i, 0
            while child[v] != -1:
                cnt += 1
                v = child[v]
            ans += (cnt+1) * a[i-1] + cnt * x

    print(ans)


if __name__ == '__main__':
    main()
