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;
const int MAXN = 502;
int N, M;
vector<int> c;
vector<int> occ[MAXN];
vector<vector<long long> > dp;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N >> M;
  c.push_back(-1);
  while (M--) {
    int x;
    cin >> x;
    x--;
    if (c.empty() || x != c.back()) {
      occ[x].push_back(c.size());
      c.push_back(x);
    }
  }
  M = c.size() - 1;
  if (M > 2 * N) {
    cout << "0\n";
    return 0;
  }
  dp.resize(M + 2, vector<long long>(M + 2, 0));
  for (int l = 1; l <= M + 1; l++) {
    for (int r = 0; r < l; r++) {
      dp[l][r] = 1;
    }
    if (l <= M && occ[c[l]].size() == 1) {
      dp[l][l] = 1;
    }
  }
  for (int sz = 2; sz <= M; sz++) {
    for (int l = 1; l + sz - 1 <= M; l++) {
      int r = l + sz - 1;
      int lm = l, rm = l;
      for (int i = l; i <= r; i++) {
        if (c[i] < c[lm]) {
          lm = rm = i;
        } else if (c[i] == c[lm]) {
          rm = i;
        }
      }
      const auto &occs = occ[c[lm]];
      if (occs[0] != lm || occs.back() != rm) {
        dp[l][r] = 0;
        continue;
      }
      long long left = 0;
      for (int a = l; a <= lm; a++) {
        left += dp[l][a - 1] * dp[a][lm - 1] % MOD;
        left %= MOD;
      }
      long long right = 0;
      for (int b = rm; b <= r; b++) {
        right += 1LL * dp[rm + 1][b] * dp[b + 1][r] % MOD;
        right %= MOD;
      }
      dp[l][r] = left * right % MOD;
      for (int i = 0; i < occs.size() - 1; i++) {
        dp[l][r] *= dp[occs[i] + 1][occs[i + 1] - 1];
        dp[l][r] %= MOD;
      }
    }
  }
  cout << dp[1][M] << '\n';
  return 0;
}