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>
#pragma GCC optimize("O3")
using namespace std;
const int mod = 998244353;
int sum(int a, int b) {
  int s = a + b;
  if (s >= mod) s -= mod;
  return s;
}
int mult(int a, int b) { return (1LL * a * b) % mod; }
const int maxN = (int)1e6 + 100;
const int maxC = 505;
int c[maxN];
int n, m;
const int LG = 21;
vector<int> all[505];
int mn[LG][maxN];
int last[maxC], first[maxC];
int lg[maxN];
int get_min(int l, int r) {
  int len = r - l + 1;
  int p = lg[len];
  return min(mn[p][l], mn[p][r - (1 << p) + 1]);
}
int dp[maxC][maxC];
int who[maxC];
int val[maxC];
int getR[maxC][maxC];
int getL[maxC][maxC];
bool is_valid[maxC][maxC];
int g[maxC];
int id[maxC];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    cin >> c[i];
    all[c[i]].push_back(i);
  }
  all[0].push_back(0);
  all[0].push_back(m + 1);
  c[0] = c[m + 1] = 0;
  for (int i = 0; i <= n; i++) {
    first[i] = all[i][0];
    last[i] = all[i].back();
  }
  memset(dp, 0, sizeof dp);
  bool bad = false;
  vector<pair<int, int> > by_first;
  for (int i = 1; i <= n; i++) {
    by_first.emplace_back(first[i], i);
  }
  for (int clr2 = 1; clr2 <= n; clr2++) {
    if (bad) break;
    for (int i = first[clr2]; i <= last[clr2]; i++) {
      if (c[i] < clr2) {
        bad = true;
        break;
      }
    }
  }
  if (bad) {
    cout << 0;
    return 0;
  }
  sort(by_first.begin(), by_first.end());
  for (int i = 0; i < n; i++) {
    who[i + 1] = by_first[i].second;
    val[i + 1] = by_first[i].first;
    id[by_first[i].second] = i + 1;
  }
  for (int i = 1; i <= n; i++) {
    getL[i][i] = first[who[i]];
    getR[i][i] = last[who[i]];
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      getL[i][j] = min(getL[i][j - 1], getL[j][j]);
      getR[i][j] = max(getR[i][j - 1], getR[j][j]);
    }
  }
  for (int len = 0; len <= n; len++) {
    for (int l = 1; l + len - 1 <= n; l++) {
      int r = l + len - 1;
      if (l > r) {
        dp[l][r] = 1;
        is_valid[l][r] = true;
        continue;
      }
      int mn = n + 1;
      bool ok = true;
      int realL = getL[l][r];
      int realR = getR[l][r];
      for (int j = l; j <= r; j++) {
        mn = min(mn, who[j]);
        if (first[who[j]] < realL || first[who[j]] > realR) {
          ok = false;
          break;
        }
      }
      is_valid[l][r] = ok;
      for (int j = 1; j < l; j++) {
        if (last[who[j]] >= realL && last[who[j]] <= realR) {
          is_valid[l][r] = false;
          break;
        }
      }
      for (int j = r + 1; j <= n; j++) {
        if (first[who[j]] <= realR) {
          is_valid[l][r] = false;
          break;
        }
      }
      if (!is_valid[l][r]) {
        dp[l][r] = 0;
        continue;
      }
      int updL = all[mn][0];
      int updR = all[mn].back();
      int coef = 1;
      if (updL != realL || updR != realR) {
        coef = mult(coef, g[mn]);
      } else {
        int nowl = l + 1;
        int nowr = nowl - 1;
        g[mn] = 1;
        for (int k = 0; k + 1 < all[mn].size(); k++) {
          while (nowr + 1 <= n && getR[nowl][nowr + 1] <= all[mn][k + 1])
            nowr++;
          g[mn] = mult(g[mn], dp[nowl][nowr]);
          nowl = nowr + 1;
          nowr = nowl - 1;
        }
        coef = g[mn];
      }
      int where = id[mn];
      int le = 0;
      int ri = 0;
      for (int x = l - 1; x <= where - 1; x++) {
        le = sum(le, mult(dp[l][x], dp[x + 1][where - 1]));
      }
      where = -1;
      for (int j = l; j <= r; j++) {
        if (val[j] > last[mn]) {
          where = j;
          break;
        }
      }
      if (where == -1)
        ri = 1;
      else {
        for (int x = where - 1; x <= r; x++) {
          ri = sum(ri, mult(dp[where][x], dp[x + 1][r]));
        }
      }
      dp[l][r] = mult(coef, mult(le, ri));
    }
  }
  cout << dp[1][n];
  return 0;
}