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;
inline void add(int &a, int b) { (a += b) >= mod && (a -= mod); }
inline int mul(int a, int b) { return 1LL * a * b - 1LL * a * b / mod * mod; }
int n, m;
int c[1000005];
int a[1000005];
vector<int> cols;
vector<int> ind[505];
int dp[1005][1005];
int dfs(int l, int r) {
  if (l > r) return 1;
  if (~dp[l][r]) return dp[l][r];
  int &res = dp[l][r];
  res = 0;
  int mn = *min_element(a + l, a + r + 1);
  int lpos = ind[mn][0], rpos = ind[mn].back();
  if (l > lpos || r < rpos) return 0;
  int left = 0, right = 0;
  for (int i = l; i <= lpos; ++i)
    add(left, mul(dfs(l, i - 1), dfs(i, lpos - 1)));
  for (int i = rpos; i <= r; ++i)
    add(right, mul(dfs(rpos + 1, i), dfs(i + 1, r)));
  res = mul(left, right);
  for (int i = 0; i < (int)ind[mn].size() - 1; ++i)
    res = mul(res, dfs(ind[mn][i] + 1, ind[mn][i + 1] - 1));
  return res;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; ++i) {
    scanf("%d", c + i);
    if (cols.empty() || cols.back() != c[i]) cols.push_back(c[i]);
  }
  m = (int)cols.size();
  if (m >= 2 * n) return puts("0") & 0;
  for (int i = 0; i < m; ++i) a[i + 1] = cols[i];
  for (int i = 1; i <= m; ++i) ind[a[i]].push_back(i);
  memset(dp, -1, sizeof(dp));
  printf("%d\n", dfs(1, m));
  return 0;
}