Problem:
Score : 1000 points

Problem Statement

You are given a permutation p = (p_1, \ldots, p_N) of \{ 1, \ldots, N \}.
You can perform the following two kinds of operations repeatedly in any order:

Pay a cost A. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the left by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_{l + 1}, p_{l + 2}, \ldots, p_r, p_l, respectively.

Pay a cost B. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the right by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_r, p_l, \ldots, p_{r - 2}, p_{r - 1}, respectively.

Find the minimum total cost required to sort p in ascending order.

Constraints

All values in input are integers.

1 \leq N \leq 5000

1 \leq A, B \leq 10^9

(p_1 \ldots, p_N) is a permutation of \{ 1, \ldots, N \}.

Input

Input is given from Standard Input in the following format:

N A B
p_1 \cdots p_N

Output

Print the minimum total cost required to sort p in ascending order.

Sample Input 1

3 20 30
3 1 2

Sample Output 1

20

Shifting (p_1, p_2, p_3) to the left by one results in p = (1, 2, 3).

Sample Input 2

4 20 30
4 2 3 1

Sample Output 2

50

One possible sequence of operations is as follows:

Shift (p_1, p_2, p_3, p_4) to the left by one. Now we have p = (2, 3, 1, 4).

Shift (p_1, p_2, p_3) to the right by one. Now we have p = (1, 2, 3, 4).

Here, the total cost is 20 + 30 = 50.

Sample Input 3

1 10 10
1

Sample Output 3

0

Sample Input 4

4 1000000000 1000000000
4 3 2 1

Sample Output 4

3000000000

Sample Input 5

9 40 50
5 3 4 7 6 1 2 9 8

Sample Output 5

220

Solution in C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

/*
https://atcoder.jp/contests/agc032/submissions/4672338 を参照しました
p[i]を右に挿入することを S -> i へコストAの辺を貼ることで表現
同様に左に挿入することを i + N -> T へコストBの辺を貼ることで表現
i < j で p[i] > p[j] なら iを右に動かすか、jを左に動かさなければならないのでi -> j + N へINFを貼る
このグラフのS-T の最小カットが求めたいものとなる(らしい)
*/
#include<stdio.h>
#include<stdlib.h>
#include<stdint.h>
#include<inttypes.h>
#include<string.h>
#include<math.h>

typedef int64_t flow_type;

const flow_type flow_inf = (int64_t) 1000000000 * 5000;

typedef struct flow_edge {
  int32_t vertex;
  int32_t next;
  flow_type capacity;
} flow_edge;

typedef struct maxFlowGraph {
  flow_edge *edge;
  int32_t *start;
  int32_t vertex_num;
  int32_t pointer;
  int32_t edge_length;
} graph;

graph* new_graph (const int32_t vertex_num) {
  graph *g = (graph *) calloc (1, sizeof (graph));
  g->vertex_num = vertex_num;
  const int32_t initial_length = 4;
  g->edge = (flow_edge *) calloc (initial_length, sizeof (flow_edge));
  g->start = (int32_t *) calloc (vertex_num, sizeof (int32_t));
  g->pointer = 0;
  g->edge_length = initial_length;
  for (int32_t i = 0; i < vertex_num; ++i) {
    g->start[i] = -1;
  }
  return g;
}

void add_edge (graph * const g, const int32_t from, const int32_t to, const flow_type capa) {
  if (g->pointer == g->edge_length) {
    g->edge_length *= 2;
    g->edge = (flow_edge *) realloc (g->edge, sizeof (flow_edge) * g->edge_length);
  }
  const int32_t p = g->pointer;
  g->edge[p] = (flow_edge) {to, g->start[from], capa};
  g->start[from] = p;
  g->edge[p + 1] = (flow_edge) {from, g->start[to], 0};
  g->start[to] = p + 1;
  g->pointer += 2;
}

flow_type dinic_dfs (const int32_t v, graph * const g, const int32_t dst, const int32_t * const level, int32_t * const iter, flow_type e) {
  if (v == dst) return e;
  flow_type sum = 0;
  for (int32_t p = iter[v]; p != -1; p = g->edge[p].next, iter[v] = p) {
    const int32_t u = g->edge[p].vertex;
    const flow_type capa = g->edge[p].capacity;
    if (level[u] <= level[v] || capa <= 0) continue;
    const flow_type f = dinic_dfs (u, g, dst, level, iter, capa < e ? capa : e);
    if (f > 0) {
      g->edge[p].capacity -= f;
      g->edge[p ^ 1].capacity += f;
      sum += f;
      e -= f;
      if (e <= 0) return sum;
    }
  }
  return sum;
}

flow_type dinic (graph * const g, const int32_t src, const int32_t dst) {
  const int32_t vertex_num = g->vertex_num;
  int32_t * const level = (int32_t *) calloc (vertex_num, sizeof (int32_t));
  int32_t * const queue = (int32_t *) calloc (vertex_num, sizeof (int32_t));
  int32_t * const iter  = (int32_t *) calloc (vertex_num, sizeof (int32_t));
  flow_type flow = 0;
  while (1) {
    memset (level, 0, sizeof (int32_t) * vertex_num);
    int32_t front = 0;
    int32_t last = 0;
    level[dst] = vertex_num;
    queue[last++] = dst;
    while (front < last && level[src] == 0) {
      const int32_t v = queue[front++];
      for (int32_t p = g->start[v]; p!=-1; p = g->edge[p].next) {
        const int32_t u = g->edge[p].vertex;
        if (g->edge[p ^ 1].capacity > 0 && level[u] == 0) {
          level[u] = level[v] - 1;
          queue[last++] = u;
        }
      }
    }
    if (level[src] == 0) break;
    memcpy (iter, g->start, sizeof (int32_t) * vertex_num);
    while (1) {
      flow_type f = dinic_dfs (src, g, dst, level, iter, flow_inf);
      if (f <= 0) break;
      flow += f;
    }
  }
  free (level);
  free (queue);
  free (iter);
  return flow;
}

typedef int32_t i32;
typedef int64_t i64;

void run (void) {
  i32 n, a, b;
  scanf ("%" SCNi32 "%" SCNi32 "%" SCNi32, &n, &a, &b);
  i32 *p = (i32 *) calloc (n, sizeof (i32));
  for (i32 i = 0; i < n; ++i) {
    scanf ("%" SCNi32, p + i);
    p[i]--;
  }
  graph *g = new_graph (2 * n + 2);
  const i32 src = 2 * n;
  const i32 dst = 2 * n + 1;
  for (i32 i = 0; i < n; ++i) {
    add_edge (g, src, i, a);
    add_edge (g, i + n, dst, b);
  }
  for (i32 i = 0; i < n; ++i) {
    for (i32 j = i + 1; j < n; ++j) {
      if (p[i] > p[j]) {
	add_edge (g, i, j + n, flow_inf);
      }
    }
  }
  i64 ans = dinic (g, src, dst);
  printf("%" PRIi64 "\n", ans);
}

int main (void) {
  run ();
  return 0;
}