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>
const long long MOD = 998244353;
using pii = std::pair<long long, long long>;
using namespace std;
const long long maxn = 505;
int32_t n, m, c[maxn], pos[maxn];
vector<int32_t> dp[maxn][maxn];
long long solve(long long step, long long l, long long r) {
  if (l > r || step == n) return 1;
  if (dp[step][l][r - l] == -1) {
    long long req = pos[step];
    if (req < l || req > r)
      dp[step][l][r - l] = solve(step + 1, l, r);
    else {
      dp[step][l][r - l] = 0;
      long long rp = 0;
      for (long long j = r; j >= req; j--) {
        rp += (solve(step + 1, req + 1, j) * solve(step + 1, j + 1, r)) % MOD;
        rp %= MOD;
      }
      for (long long i = l; i <= req; i++) {
        long long part = solve(step + 1, l, i - 1);
        part *= solve(step + 1, i, req - 1);
        part %= MOD;
        part *= rp;
        part %= MOD;
        dp[step][l][r - l] += part;
        dp[step][l][r - l] %= MOD;
      }
    }
  }
  return dp[step][l][r - l];
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (long long i = 0; i < n; i++) {
    cin >> c[i];
    c[i]--;
    pos[c[i]] = i;
  }
  for (long long i = 0; i < n; i++) {
    long long take = -1;
    if (i > 0) take = pos[i - 1];
    for (long long j = 0; j < n; j++) {
      if (j < take) {
        dp[i][j].resize(take - j + 1);
        for (long long k = j; k <= take; k++) dp[i][j][k - j] = -1;
      } else {
        dp[i][j].resize(n - j);
        for (long long k = j; k < n; k++) dp[i][j][k - j] = -1;
      }
    }
  }
  cout << solve(0, 0, n - 1) << "\n";
  return 0;
}