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>
#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  return s * w;
}
long long f[1003][1003];
bool vis[1003][1003];
int a[1000003];
vector<int> v[1000003];
long long dfs(int l, int r) {
  if (l > r) return 1;
  if (vis[l][r]) return f[l][r];
  vis[l][r] = 1;
  long long ans1 = 0, ans2 = 0;
  int now = l;
  for (int i = l; i <= r; i++)
    if (a[i] < a[now]) now = i;
  int p = v[a[now]][0], q = v[a[now]].back();
  for (int i = l; i <= p; i++)
    ans1 += dfs(l, i - 1) * dfs(i, p - 1) % 998244353LL, ans1 %= 998244353LL;
  for (int i = q; i <= r; i++)
    ans2 += dfs(q + 1, i) * dfs(i + 1, r) % 998244353LL, ans2 %= 998244353LL;
  long long ans = ans1 * ans2 % 998244353LL;
  for (int i = 0; i + 1 < v[a[now]].size(); i++)
    ans *= dfs(v[a[now]][i] + 1, v[a[now]][i + 1] - 1), ans %= 998244353LL;
  return f[l][r] = ans;
}
int main() {
  int m = read(), n = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  n = unique(a + 1, a + n + 1) - a - 1;
  if (n > 2 * m) return 0 & puts("0");
  for (int i = 1; i <= n; i++) v[a[i]].push_back(i);
  printf("%lld\n", dfs(1, n));
  return 0;
}
