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;
constexpr int maxn = 1005;
constexpr long long mod = 998244353;
int n, m;
vector<int> v, tmp;
vector<int> pos[maxn];
int LoID[maxn];
long long dp[maxn][maxn];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int x;
    cin >> x;
    tmp.emplace_back(x);
  }
  for (int i = 0; i < m; i++) {
    if (i > 0) {
      if (v.back() == tmp[i]) {
        continue;
      }
    }
    v.emplace_back(tmp[i]);
    pos[tmp[i]].emplace_back((int)v.size());
  }
  m = (int)v.size();
  if (m > 2 * n) {
    cout << 0 << endl;
    exit(0);
  }
  for (int i = 0; i < m; i++) {
    LoID[i + 1] = v[i];
  }
  for (int i = 1; i <= m + 1; i++)
    for (int j = 1; j <= m + 1; j++) dp[i][j] = 1;
  for (int len = 0; len <= m; len++) {
    for (int i = 1; i + len <= m; i++) {
      int mnid = 999;
      for (int j = i; j <= i + len; j++) mnid = min(mnid, LoID[j]);
      int mn = m + 1, mx = 0;
      for (auto x : pos[mnid]) {
        mn = min(mn, x);
        mx = max(mx, x);
      }
      if (mn < i || mx > i + len) {
        dp[i][i + len] = 0;
        continue;
      }
      long long sumleft = (i != mn) ? 2 * dp[i][mn - 1] % mod : 1;
      for (int j = i; j < mn - 1; j++) {
        sumleft = (sumleft + dp[i][j] * dp[j + 1][mn - 1] % mod) % mod;
      }
      long long sumright = (mx != i + len) ? 2 * dp[mx + 1][i + len] % mod : 1;
      for (int j = mx + 1; j < i + len; j++) {
        sumright = (sumright + dp[mx + 1][j] * dp[j + 1][i + len] % mod) % mod;
      }
      long long res = sumleft * sumright % mod;
      for (int i = 1; i < (int)pos[mnid].size(); i++) {
        res = res * dp[pos[mnid][i - 1] + 1][pos[mnid][i] - 1] % mod;
      }
      dp[i][i + len] = res;
    }
  }
  cout << dp[1][m] << endl;
  return 0;
}