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;
struct kek {
  long long l, r, col;
  bool operator<(const kek& a) const { return (l < a.l); }
};
long long mod = 998244353;
vector<long long> c;
vector<vector<long long>> dp;
vector<pair<long long, long long>> otr;
vector<vector<long long>> maxx;
vector<pair<long long, long long>> bounds;
vector<kek> z;
long long n, m;
long long mx(long long l, long long r) {
  if (r < l) return -1;
  if (maxx[l][r] != -1) return maxx[l][r];
  for (long long i = l; i <= r; i++) {
    maxx[l][r] = max(maxx[l][r], z[i].r);
  }
  return maxx[l][r];
}
long long calc(long long l, long long r) {
  if (r < l) return 1;
  if (dp[l][r] != -1) return dp[l][r];
  long long pref = 0, suf = 0, pos = l;
  for (long long i = l; i <= r; i++) {
    if (z[i].col < z[pos].col) pos = i;
  }
  for (long long j = pos; j >= l; j--) {
    if (mx(l, j - 1) < z[j].l && z[j].l >= bounds[z[pos].col].first) {
      pref = (pref + calc(l, j - 1) * calc(j, pos - 1)) % mod;
    }
  }
  long long maxx = -1;
  for (long long j = pos; j <= r; j++) {
    maxx = max(maxx, z[j].r);
    if (j != r && z[j + 1].l <= maxx) continue;
    if (maxx > bounds[z[pos].col].second) continue;
    suf = (suf + calc(j + 1, r) * calc(pos + 1, j)) % mod;
  }
  dp[l][r] = (pref * suf) % mod;
  return dp[l][r];
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  c = vector<long long>(m);
  dp = vector<vector<long long>>(n, vector<long long>(n, -1));
  maxx = vector<vector<long long>>(n, vector<long long>(n, -1));
  otr = vector<pair<long long, long long>>(n, {m, -1});
  bounds = vector<pair<long long, long long>>(n, {0, m - 1});
  vector<long long> pos(n, -1);
  for (long long i = 0; i < m; i++) {
    cin >> c[i];
    c[i]--;
    otr[c[i]].first = min(otr[c[i]].first, i);
    otr[c[i]].second = max(otr[c[i]].second, i);
    pos[c[i]] = i;
    for (long long j = 0; j < c[i]; j++) {
      bounds[c[i]].first = max(bounds[c[i]].first, pos[j] + 1);
    }
  }
  pos = vector<long long>(n, m);
  for (long long i = m - 1; i >= 0; i--) {
    pos[c[i]] = i;
    for (long long j = 0; j < c[i]; j++) {
      bounds[c[i]].second = min(bounds[c[i]].second, pos[j] - 1);
    }
  }
  for (long long i = 0; i < n; i++) {
    z.push_back({otr[i].first, otr[i].second, i});
    for (long long j = 0; j < n; j++) {
      if (i == j) continue;
      long long l = max(otr[i].first, otr[j].first),
                r = min(otr[i].second, otr[j].second);
      if (r < l) continue;
      if (l == otr[i].first && r == otr[i].second && i > j) continue;
      if (l == otr[j].first && r == otr[j].second && j > i) continue;
      cout << 0;
      return 0;
    }
  }
  sort(z.begin(), z.end());
  cout << calc(0, n - 1);
}