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 N = 1000 + 1, M = 1e6 + 1, mod = 998244353;
int n, m;
int c[M];
int cnt[N];
int sz;
int a[N];
int in[N], out[N];
bool is_opened[N];
long long mp[N][N];
long long solve(int l, int r) {
  if (l == r) return 1;
  if (mp[l][r] != -1) return mp[l][r];
  int m = -1, mn = N;
  for (int i = l; i < r; ++i)
    if (a[i] < mn) {
      m = i;
      mn = a[i];
    }
  if (mn == N) return 1;
  long long resl = 0, resr = 0;
  stack<int> opened;
  int near = l;
  for (int i = l; i <= m; ++i) {
    while (!opened.empty() && out[opened.top()] == i) opened.pop();
    if (opened.empty())
      if (a[i] == N) near = i + 1;
    if (a[i] != N) opened.push(a[i]);
  }
  while (!opened.empty()) opened.pop();
  for (int i = near; i <= m; ++i) {
    while (!opened.empty() && out[opened.top()] == i) opened.pop();
    if (opened.empty()) resl = (resl + solve(l, i) * solve(i, m)) % mod;
    if (a[i] != N) opened.push(a[i]);
  }
  while (!opened.empty()) opened.pop();
  for (int i = out[a[m]]; i <= r; ++i) {
    while (!opened.empty() && out[opened.top()] == i) opened.pop();
    if (opened.empty()) {
      resr = (resr + solve(out[a[m]], i) * solve(i, r)) % mod;
      if (a[i] == N) break;
    }
    if (a[i] != N) opened.push(a[i]);
  }
  int resm = solve(m + 1, out[a[m]]);
  return mp[l][r] = resl * resr % mod * resm % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < 2 * n; ++i)
    for (int j = i + 1; j <= 2 * n; ++j) mp[i][j] = -1;
  for (int i = 0; i < m; ++i) {
    cin >> c[i];
    ++cnt[c[i]];
  }
  stack<int> opened;
  bool bar = false;
  int last = -1;
  for (int i = 0; i < m; ++i) {
    --cnt[c[i]];
    if (i != m - 1 && c[i + 1] == c[i]) continue;
    if (is_opened[c[i]]) {
      if (!opened.empty() && opened.top() != c[i]) {
        cout << "0\n";
        return 0;
      }
      if (cnt[c[i]] == 0) {
        out[c[i]] = sz;
        opened.pop();
        is_opened[c[i]] = false;
      } else if (!bar && c[i] != last) {
        a[sz++] = N;
        bar = true;
      }
    } else {
      if (!opened.empty() && c[i] < opened.top()) {
        cout << "0\n";
        return 0;
      }
      bar = false;
      in[c[i]] = sz;
      a[sz++] = c[i];
      opened.push(c[i]);
      is_opened[c[i]] = true;
      if (cnt[c[i]] == 0) {
        out[c[i]] = sz;
        opened.pop();
        is_opened[c[i]] = false;
      }
    }
    last = c[i];
  }
  cout << solve(0, sz) << '\n';
}