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 int MOD = 998244353;
int n, m;
vector<int> c;
long long int dp[1010][1010];
int mn[510], mx[510];
long long int solve(int l, int r) {
  if (l > r) return 1;
  if (dp[l][r] >= 0) return dp[l][r];
  if (l == r) {
    if (mn[c[l]] == l && mx[c[l]] == l)
      return dp[l][r] = 1;
    else
      return dp[l][r] = 0;
  }
  int mn0 = *min_element(c.begin() + l, c.begin() + r + 1);
  vector<int> ind;
  for (int i = l; i <= r; i++)
    if (c[i] == mn0) ind.push_back(i);
  int x = ind[0], y = ind.back();
  if (mn[mn0] != x || mx[mn0] != y) return dp[l][r] = 0;
  long long int s1 = 0, s2 = 0;
  for (int i = l; i <= x; i++) (s1 += solve(l, i - 1) * solve(i, x - 1)) %= MOD;
  for (int i = y; i <= r; i++) (s2 += solve(y + 1, i) * solve(i + 1, r)) %= MOD;
  long long int ret = s1 * s2 % MOD;
  for (int i = 0; i < (int)ind.size() - 1; i++)
    (ret *= solve(ind[i] + 1, ind[i + 1] - 1)) %= MOD;
  return dp[l][r] = ret;
}
int main() {
  scanf("%d %d", &n, &m);
  c.resize(m);
  for (int i = 0; i < m; i++) {
    scanf("%d", &c[i]);
  }
  c.erase(unique(c.begin(), c.end()), c.end());
  m = c.size();
  if (m > 2 * n + 1) {
    cout << 0 << endl;
    return 0;
  }
  for (int i = 0; i < m; i++)
    for (int j = i; j < m; j++) dp[i][j] = -1;
  fill(mn, mn + n + 1, -1);
  for (int i = 0; i < m; i++) {
    if (mn[c[i]] == -1) mn[c[i]] = i;
    mx[c[i]] = i;
  }
  cout << solve(0, m - 1) << endl;
  return 0;
}