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 mod = 998244353;
int a[509], pos[509];
int dp[509][509];
int mn[509][509];
int ans, n, m;
int main() {
  memset(dp, 0, sizeof(dp));
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i], pos[a[i]] = i;
  for (int i = 1; i <= n; i++) dp[i][i] = 1, mn[i][i] = i;
  for (int k = 1; k < n; k++)
    for (int i = 1; i + k <= n; i++) {
      if (a[mn[i][i + k - 1]] > a[i + k])
        mn[i][i + k] = i + k;
      else
        mn[i][i + k] = mn[i][i + k - 1];
      int j = i + k, mid = mn[i][i + k];
      long long sum1 = 0, ans = 1, sum2 = 0;
      for (int l = i; l <= mid; l++) {
        ans = 1;
        if (i < l - 1) ans = (ans * dp[i][l - 1]) % mod;
        if (l < mid - 1) ans = (ans * dp[l][mid - 1]) % mod;
        sum1 += ans;
        if (sum1 >= mod) sum1 -= mod;
      }
      for (int r = mid; r <= j; r++) {
        ans = 1;
        if (mid + 1 < r) ans = (ans * dp[mid + 1][r]) % mod;
        if (r + 1 < j) ans = (ans * dp[r + 1][j]) % mod;
        sum2 += ans;
        if (sum2 >= mod) sum2 -= mod;
      }
      dp[i][j] = sum1 * sum2 % mod;
    }
  cout << dp[1][n] << endl;
  return 0;
}