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;
int a[511];
int rea[511];
long long dp1[511][511];
long long dp2[511][511];
int mn[511][511];
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d", &a[i]);
    rea[a[i]] = i;
  }
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) mn[i][j] = n;
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++)
      for (int k = i; k <= j; k++) mn[i][j] = min(mn[i][j], a[k]);
  for (int i = 1; i <= n; i++) dp1[i][rea[i]] = dp2[i][rea[i]] = 1;
  for (int t = n; t >= 1; t--) {
    int p = rea[t];
    for (int i = p; i >= 1 && mn[i][p] == t; i--)
      for (int j = p; j >= i; j--) {
        if (i != p && j == p)
          dp1[t][i] =
              (dp1[t][i] + dp1[mn[i][p - 1]][i] * dp2[mn[i][p - 1]][p - 1]) %
              998244353;
        if (i != p && j != p)
          if (i != j) {
            long long plu =
                (dp1[mn[i][j - 1]][i] * dp2[mn[i][j - 1]][j - 1]) % 998244353;
            plu = (plu * dp1[mn[j][p - 1]][j]) % 998244353;
            plu = (plu * dp2[mn[j][p - 1]][p - 1]) % 998244353;
            dp1[t][i] = (dp1[t][i] + plu) % 998244353;
          }
        if (i != p && j != p)
          if (i == j)
            dp1[t][i] =
                (dp1[t][i] + dp1[mn[i][p - 1]][i] * dp2[mn[i][p - 1]][p - 1]) %
                998244353;
      }
    for (int i = p; i <= n && mn[p][i] == t; i++)
      for (int j = p; j <= i; j++) {
        if (i != p && j == p)
          dp2[t][i] =
              (dp2[t][i] + dp1[mn[p + 1][i]][p + 1] * dp2[mn[p + 1][i]][i]) %
              998244353;
        if (i != p && j != p)
          if (i != j) {
            long long plu =
                (dp1[mn[j + 1][i]][j + 1] * dp2[mn[j + 1][i]][i]) % 998244353;
            plu = (plu * dp1[mn[p + 1][j]][p + 1]) % 998244353;
            plu = (plu * dp2[mn[p + 1][j]][j]) % 998244353;
            dp2[t][i] = (dp2[t][i] + plu) % 998244353;
          }
        if (i != p && j != p)
          if (i == j)
            dp2[t][i] =
                (dp2[t][i] + dp1[mn[p + 1][i]][p + 1] * dp2[mn[p + 1][i]][i]) %
                998244353;
      }
  }
  printf("%lld\n", (dp1[1][1] * dp2[1][n]) % 998244353);
}