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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<int> d(m);
  for (int& di : d) cin >> di;
  vector<int> c;
  vector<vector<int> > pos(n + 1);
  for (int di : d) {
    if (c.empty() or c.back() != di) {
      pos[di].push_back(int(c.size()));
      c.push_back(di);
    }
  }
  if (int(c.size()) > 2 * n) {
    cout << 0 << '\n';
    return (0);
  }
  n = int(c.size());
  const int mod = 998244353;
  vector<vector<int> > dp(n, vector<int>(n, -1));
  auto add = [&](long long x, long long y) { return (x + y) % mod; };
  auto mul = [&](long long x, long long y) { return (x * y) % mod; };
  function<int(int, int)> rec = [&](int l, int r) -> int {
    if (l > r) return 1;
    if (l == r) return (l == pos[c[l]].front() and pos[c[l]].back() == r);
    if (dp[l][r] != -1) return dp[l][r];
    int mn = c[l];
    for (int i = l; i <= r; i++) mn = min(mn, c[i]);
    int x = pos[mn].front();
    int y = pos[mn].back();
    if (not(l <= x and y <= r)) return dp[l][r] = 0;
    int left = 0;
    int right = 0;
    for (int a = l; a <= x; a++) {
      left = add(left, mul(rec(l, a - 1), rec(a, x - 1)));
    }
    for (int b = y; b <= r; b++) {
      right = add(right, mul(rec(y + 1, b), rec(b + 1, r)));
    }
    int ret = mul(left, right);
    for (int i = 0; i + 1 < int(pos[mn].size()); i++) {
      int p1 = pos[mn][i];
      int p2 = pos[mn][i + 1];
      ret = mul(ret, rec(p1 + 1, p2 - 1));
    }
    return dp[l][r] = ret;
  };
  cout << rec(0, n - 1) << '\n';
  return (0);
}