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 long long maxn = 5e2 + 10, mod = 998244353;
long long a[maxn], n, m, s[maxn][maxn], f[maxn][maxn], mark[maxn];
void read() {
  cin >> n >> m;
  for (int i = 1; i <= m; i++) cin >> a[i], mark[a[i]] = i;
}
void add(long long &t1, long long t2) {
  t1 += t2;
  if (t1 >= mod) t1 -= mod;
}
long long multi(long long t1, long long t2) { return (t1 * t2) % mod; }
void process() {
  for (int i = 1; i <= m; i++) {
    s[i][i] = a[i];
    for (int j = i + 1; j <= m; j++) s[i][j] = min(s[i][j - 1], a[j]);
    for (int j = i; j <= m; j++) s[i][j] = mark[s[i][j]];
  }
  for (int i = n; i; i--) {
    int vt = mark[i];
    int l = vt - 1, r = vt + 1;
    f[vt][vt] = 1;
    while (a[l] > a[vt]) {
      int t = s[l][vt - 1];
      add(f[vt][l], multi(f[t][l], f[t][vt - 1]));
      add(f[vt][l], multi(f[t][l], f[t][vt - 1]));
      for (int j = vt - 1; j > l; j--) {
        int t1 = s[j][vt - 1], t2 = s[l][j - 1];
        add(f[vt][l], multi(multi(f[t1][j], f[t1][vt - 1]),
                            multi(f[t2][l], f[t2][j - 1])));
      }
      l--;
    }
    while (a[r] > a[vt]) {
      int t = s[vt + 1][r];
      add(f[vt][r], multi(f[t][vt + 1], f[t][r]));
      add(f[vt][r], multi(f[t][vt + 1], f[t][r]));
      for (int j = vt + 1; j < r; j++) {
        int t1 = s[vt + 1][j], t2 = s[j + 1][r];
        add(f[vt][r], multi(multi(f[t1][vt + 1], f[t1][j]),
                            multi(f[t2][j + 1], f[t2][r])));
      }
      r++;
    }
  }
  cout << multi(f[mark[1]][1], f[mark[1]][m]);
}
int main() {
  read();
  process();
  return 0;
}