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;
using lli = long long;
using pii = pair<int, int>;
const int MOD = 998244353;
int n, m, arr[1000000], li[501], ri[501], dp[1000][1000];
int lm[501], rm[501], lp[501], rp[501];
vector<pii> adj;
int DP(int l, int r) {
  if (l >= r) return 1;
  int &ret = dp[l][r];
  if (ret != -1) return ret;
  int mx = n;
  for (int i = l; i <= r; i++) mx = min(mx, adj[i].first);
  int mv = DP(li[mx] + 1, ri[mx] - 1), lv = 0, rv = 0, cnt = 0;
  cnt = 1;
  for (int j = li[mx]; j >= l && lp[adj[j].first] > lm[mx]; j--) {
    cnt += adj[j].second;
    if (!cnt && adj[j].second == -1)
      lv = (lv + 1ll * DP(j, li[mx] - 1) * DP(l, j - 1) % MOD) % MOD;
  }
  cnt = -1;
  for (int j = ri[mx]; j <= r && rp[adj[j].first] < rm[mx]; j++) {
    cnt += adj[j].second;
    if (!cnt && adj[j].second == 1)
      rv = (rv + 1ll * DP(ri[mx] + 1, j) * DP(j + 1, r) % MOD) % MOD;
  }
  return ret = 1ll * lv * mv % MOD * rv % MOD;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) scanf("%d", arr + i);
  memset(li, -1, sizeof(li));
  memset(ri, -1, sizeof(ri));
  for (int i = 0; i < m; i++)
    if (li[arr[i]] == -1) li[arr[i]] = i;
  for (int i = m - 1; i >= 0; i--)
    if (ri[arr[i]] == -1) ri[arr[i]] = i;
  for (int i = 1; i <= n; i++) lm[i] = lp[i] = li[i], rm[i] = rp[i] = ri[i];
  vector<pii> tmp;
  for (int i = 1; i <= n; i++) {
    tmp.push_back({li[i], i});
    tmp.push_back({ri[i] + 1, -i});
  }
  sort(tmp.begin(), tmp.end());
  set<int> tr;
  for (int i = 0, j = 0; i < m; i++) {
    while (j < (int)tmp.size() && tmp[j].first <= i) {
      if (tmp[j].second < 0)
        tr.erase(-tmp[j].second);
      else
        tr.insert(tmp[j].second);
      j++;
    }
    if (tr.upper_bound(arr[i]) != tr.end()) {
      puts("0");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int &j = lm[i]; j >= 0 && arr[j] >= i; j--)
      ;
    for (int &j = rm[i]; j < m && arr[j] >= i; j++)
      ;
  }
  for (int i = 1; i <= n; i++) {
    adj.push_back({i, -1});
    adj.push_back({i, 1});
  }
  sort(adj.begin(), adj.end(), [&](const pii &a, const pii &b) {
    int ai = a.second == -1 ? li[a.first] : ri[a.first];
    int bi = b.second == -1 ? li[b.first] : ri[b.first];
    return ai == bi ? a.second < b.second : ai < bi;
  });
  for (int i = 0; i < n * 2; i++) {
    if (adj[i].second == -1)
      li[adj[i].first] = i;
    else
      ri[adj[i].first] = i;
  }
  memset(dp, -1, sizeof(dp));
  printf("%d\n", DP(0, n * 2 - 1));
  return 0;
}