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 long long INF = (long long)1e18;
const long long mod = 998244353;
const int inf = (int)1e9;
const double eps = (double)1e-9;
const double pi = acos(-1.0);
const int dx[] = {0, 0, 1, 0, -1};
const int dy[] = {0, 1, 0, -1, 0};
const int N = 2000500;
int n, m, a[N], lv[N], rv[N];
pair<int, int> t[4 * N];
vector<int> v[N];
long long dp[511][511];
bool used[511][511];
void build(int vv, int tl, int tr) {
  if (tl == tr) {
    t[vv].first = lv[a[tl]];
    t[vv].second = rv[a[tl]];
    return;
  }
  int tm = (tl + tr) >> 1;
  build(vv + vv, tl, tm);
  build(vv + vv + 1, tm + 1, tr);
  t[vv].first = min(t[vv + vv].first, t[vv + vv + 1].first);
  t[vv].second = max(t[vv + vv].second, t[vv + vv + 1].second);
}
pair<int, int> get(int vv, int tl, int tr, int l, int r) {
  if (l > r) return make_pair(m + 1, -1);
  if (tl == l && tr == r) return t[vv];
  int tm = (tl + tr) >> 1;
  pair<int, int> prl = get(vv + vv, tl, tm, l, min(r, tm));
  pair<int, int> prr = get(vv + vv + 1, tm + 1, tr, max(l, tm + 1), r);
  return make_pair(min(prl.first, prr.first), max(prl.second, prr.second));
}
long long rec(int La, int Ra) {
  int lc = a[La], rc = a[Ra];
  if (!lc || !rc) return 1;
  if (La > Ra) return 1;
  if (used[lc][rc]) return dp[lc][rc];
  used[lc][rc] = 1;
  int mx = lc;
  for (int i = 1; i <= n; ++i) {
    if (La <= lv[i] && rv[i] <= Ra && i <= mx) mx = i;
  }
  long long answer = 1;
  for (int i = 0; i + 1 < (int)v[mx].size(); ++i) {
    answer = (answer * rec(v[mx][i] + 1, v[mx][i + 1] - 1)) % mod;
  }
  long long ansl = rec(La, lv[mx] - 1);
  int la = lv[mx];
  while (La < la) {
    la = lv[a[la - 1]];
    ansl += rec(La, la - 1) * rec(la, lv[mx] - 1);
    ansl %= mod;
  }
  long long ansr = rec(rv[mx] + 1, Ra);
  int ra = rv[mx];
  while (ra < Ra) {
    ra = rv[a[ra + 1]];
    ansr += rec(rv[mx] + 1, ra) * rec(ra + 1, Ra);
    ansr %= mod;
  }
  dp[lc][rc] = answer * ansl % mod * ansr % mod;
  return dp[lc][rc];
}
int main() {
  cin.tie(NULL);
  cout.tie(NULL);
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 1; i <= m; ++i) {
    cin >> a[i];
  }
  int M = 1;
  for (int i = 2; i <= m; ++i)
    if (a[i] != a[M]) {
      M++;
      a[M] = a[i];
    }
  m = M;
  for (int i = 1; i <= m; ++i) {
    if (lv[a[i]] == 0) lv[a[i]] = i;
    rv[a[i]] = i;
    v[a[i]].push_back(i);
  }
  for (int i = 1; i <= n; ++i)
    for (int j = i + 1; j <= n; ++j)
      if (lv[j] < lv[i] && rv[i] < rv[j]) {
        cout << 0;
        return 0;
      }
  build(1, 1, m);
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < (int)v[i].size() - 1; ++j) {
      pair<int, int> pr = get(1, 1, m, v[i][j] + 1, v[i][j + 1] - 1);
      if (pr.first <= v[i][j] || pr.second >= v[i][j + 1]) {
        cout << 0;
        return 0;
      }
    }
  }
  cout << rec(1, m);
  return 0;
}