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;
void Freopen() {
  freopen(
      "title"
      ".in",
      "r", stdin);
  freopen(
      "title"
      ".out",
      "w", stdout);
}
int read() {
  int g = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || '9' < ch) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    g = g * 10 + ch - '0';
    ch = getchar();
  }
  return g * f;
}
const int N = 505;
const int INF = 1e9;
const int M = 1e6 + 5;
const int mod = 998244353;
int s[N << 1][N << 1], n, m, a[M], cnt;
vector<int> pos[N];
int dfs(int l, int r) {
  if (l > r) return 1;
  if (~s[l][r]) return s[l][r];
  int mi = INF;
  for (int i = (l); i <= (r); i++) mi = min(mi, a[i]);
  int L = pos[mi][0], R = pos[mi][pos[mi].size() - 1], al = 0, ar = 0, ans;
  if (L < l || R > r) return s[l][r] = 0;
  for (int i = (l); i <= (L); i++)
    al = (al + 1ll * dfs(l, i - 1) * dfs(i, L - 1) % mod) % mod;
  for (int i = (R); i <= (r); i++)
    ar = (ar + 1ll * dfs(R + 1, i) * dfs(i + 1, r) % mod) % mod;
  ans = 1ll * al * ar % mod;
  for (int i = (0); i <= ((int)pos[mi].size() - 2); i++)
    ans = 1ll * ans * dfs(pos[mi][i] + 1, pos[mi][i + 1] - 1) % mod;
  return s[l][r] = ans;
}
signed main() {
  n = read(), m = read();
  for (int i = (1); i <= (m); i++) a[i] = read();
  int cnt = 0;
  for (int i = (1); i <= (m); i++) {
    if (a[i] == a[i + 1]) continue;
    a[++cnt] = a[i];
  }
  if (cnt > 2 * n) return cout << 0, signed();
  for (int i = (1); i <= (cnt); i++) pos[a[i]].push_back(i);
  memset(s, -1, sizeof(s));
  return cout << dfs(1, cnt), signed();
}