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;
int c[1000100];
int n, m;
int st[1000100], ed[1000100];
int flag_st[1000100], flag_ed[1000100];
vector<int> res[555];
const int MM = 555;
int dp[MM][MM];
map<pair<int, int>, int> mp;
inline int add(int &u, int v) {
  u += v;
  if (u >= 998244353) u -= 998244353;
}
int calc(int L, int R) {
  if (L > R) return 1;
  if (!flag_st[L] || !flag_ed[R]) return 0;
  int &ret = dp[flag_st[L]][flag_ed[R]];
  if (~ret) return ret;
  ret = 0;
  int mn_id = 0x3f3f3f3f, max_id = 0;
  for (int i = 1; i <= n; i++) {
    if (st[i] >= L and st[i] <= R and ed[i] > R) return ret = 0;
    if (ed[i] >= L and ed[i] <= R and st[i] < L) return ret = 0;
    if (st[i] >= L and ed[i] <= R) mn_id = min(mn_id, i);
    if (st[i] < L and ed[i] > R) max_id = max(max_id, i);
  }
  if (max_id > mn_id) return ret = 0;
  int ans = 1;
  for (int i = 0; i < res[mn_id].size() - 1; i++) {
    ans = (long long)ans * calc(res[mn_id][i] + 1, res[mn_id][i + 1] - 1) %
          998244353;
    if (ans == 0) return ret = 0;
  }
  int mx = calc(L, st[mn_id] - 1);
  int my = calc(ed[mn_id] + 1, R);
  for (int i = 1; i <= n; i++)
    if (i != mn_id) {
      if (st[i] >= L and ed[i] <= st[mn_id]) {
        add(mx, (long long)calc(L, st[i] - 1) * calc(st[i], st[mn_id] - 1) %
                    998244353);
      }
      if (ed[i] <= R and ed[i] >= ed[mn_id]) {
        add(my, (long long)calc(ed[mn_id] + 1, ed[i]) * calc(ed[i] + 1, R) %
                    998244353);
      }
    }
  ans = (long long)ans * mx % 998244353 * my % 998244353;
  return ret = ans;
}
int main() {
  cin >> n >> m;
  memset(dp, -1, sizeof dp);
  for (int i = 1; i <= m; i++) scanf("%d", c + i);
  for (int i = 1; i <= m; i++) res[c[i]].push_back(i);
  for (int i = 1; i <= n; i++) st[i] = res[i][0], ed[i] = res[i].back();
  for (int i = 1; i <= n; i++) flag_st[st[i]] = i, flag_ed[ed[i]] = i;
  flag_ed[0] = 0, flag_st[m + 1] = 0;
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++) {
      if (st[i] > ed[j]) continue;
      if (ed[i] < st[j]) continue;
      if (st[j] < st[i]) return puts("0"), 0;
      if (ed[j] > ed[i]) return puts("0"), 0;
    }
  cout << calc(1, m) << endl;
}