Problem: This is the first subtask of problem F. The only differences between this and the second subtask are the constraints on the value of m and the time limit. You need to solve both subtasks in order to hack this one.

There are n+1 distinct colours in the universe, numbered 0 through n. There is a strip of paper m centimetres long initially painted with colour 0. 

Alice took a brush and painted the strip using the following process. For each i from 1 to n, in this order, she picks two integers 0 ≤ a_i < b_i ≤ m, such that the segment [a_i, b_i] is currently painted with a single colour, and repaints it with colour i. 

Alice chose the segments in such a way that each centimetre is now painted in some colour other than 0. Formally, the segment [i-1, i] is painted with colour c_i (c_i ≠ 0). Every colour other than 0 is visible on the strip.

Count the number of different pairs of sequences \\{a_i\}_{i=1}^n, \\{b_i\}_{i=1}^n that result in this configuration. 

Since this number may be large, output it modulo 998244353.

Input

The first line contains a two integers n, m (1 ≤ n ≤ 500, n = m) — the number of colours excluding the colour 0 and the length of the paper, respectively.

The second line contains m space separated integers c_1, c_2, …, c_m (1 ≤ c_i ≤ n) — the colour visible on the segment [i-1, i] after the process ends. It is guaranteed that for all j between 1 and n there is an index k such that c_k = j.

Note that since in this subtask n = m, this means that c is a permutation of integers 1 through n.

Output

Output a single integer — the number of ways Alice can perform the painting, modulo 998244353.

Examples

Input


3 3
1 2 3


Output


5


Input


7 7
4 5 1 6 2 3 7


Output


165

Note

In the first example, there are 5 ways, all depicted in the figure below. Here, 0 is white, 1 is red, 2 is green and 3 is blue.

<image>

Below is an example of a painting process that is not valid, as in the second step the segment 1 3 is not single colour, and thus may not be repainted with colour 2.

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 505;
const int MOD = 998244353;
int n, ans;
int c[N], p[N];
int dp[N][N], sm[N][N];
int dfs(int x, int y) {
  int C = sm[x][y];
  if (dp[x][y] != -1) return dp[x][y];
  if (x == y) return dp[x][y] = 1;
  if (x > y) return dp[x][y] = 0;
  dp[x][y] = 0ll;
  int lft = 2ll * dfs(x, p[C] - 1), rht = 2ll * dfs(p[C] + 1, y);
  for (int i = x + 1; i <= p[C] - 1; i++)
    lft = (1ll * lft + 1ll * dfs(i, p[C] - 1) * dfs(x, i - 1) % MOD) % MOD;
  for (int i = p[C] + 1; i <= y - 1; i++)
    rht = (1ll * rht + 1ll * dfs(p[C] + 1, i) * dfs(i + 1, y) % MOD) % MOD;
  if (!lft) lft++;
  if (!rht) rht++;
  ans = (1ll * lft * rht) % MOD;
  return dp[x][y] = ans;
}
int main() {
  for (int i = 0; i < N; i++)
    for (int j = 0; j < N; j++) dp[i][j] = -1;
  scanf("%d%d", &n, &n);
  for (int i = 1; i <= n; i++) scanf("%d", c + i);
  for (int i = 1; i <= n; i++) {
    int mn = c[i];
    p[c[i]] = i;
    for (int j = i; j <= n; j++) {
      mn = min(mn, c[j]);
      sm[i][j] = mn;
    }
  }
  printf("%d\n", dfs(1, n));
  return 0;
}