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 n, m, a[2000000], dp[1005][1005];
vector<int> v[1005];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) scanf("%d", a + i);
  m = unique(a + 1, a + m + 1) - a - 1;
  if (m > n * 2) return puts("0"), 0;
  for (int i = 1; i <= m; i++) v[a[i]].push_back(i);
  for (int i = 1; i <= m + 1; i++) dp[i][i - 1] = 1;
  for (int l = 1; l <= m; l++)
    for (int i = 1, j = l; j <= m; i++, ++j) {
      int cre = *min_element(a + i, a + j + 1);
      int kl = v[cre].front(), lk = v[cre].back();
      if (kl < i || lk > j) {
        dp[i][j] = 0;
        continue;
      }
      long long ll = 0, rr = 0;
      for (int p = i; p <= kl; p++)
        (ll += 1ll * dp[i][p - 1] * dp[p][kl - 1]) %= mod;
      for (int p = lk; p <= j; p++)
        (rr += 1ll * dp[lk + 1][p] * dp[p + 1][j]) %= mod;
      dp[i][j] = 1ll * ll * rr % mod;
      for (int p = 1; p < v[cre].size(); p++) {
        if (v[cre][p - 1] + 1 < v[cre][p])
          dp[i][j] =
              1ll * dp[i][j] * dp[v[cre][p - 1] + 1][v[cre][p] - 1] % mod;
      }
    }
  cout << dp[1][m] % mod;
}