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 int MOD = 998244353;
const int MAXN = 505;
const int MAXM = 1000005;
int n, m, c[MAXM];
vector<int> v[MAXN];
int a[MAXN];
long long dp[MAXN][MAXN];
long long calc(int L, int R) {
  if (R <= L) return 1;
  if (dp[L][R] != -1) return dp[L][R];
  int j = L;
  long long ret = 0;
  for (int i = L; i <= R; i++) {
    if (a[i] < a[j]) j = i;
    ret = (ret + calc(L, j - 1) * calc(j + 1, i) % MOD * calc(i + 1, R)) % MOD;
  }
  return dp[L][R] = ret;
}
long long solve(int L, int R) {
  if (R < L) return 1;
  long long ret = 1;
  int k = 0;
  for (int i = L; i <= R;) {
    for (int j = 0; j < (int)v[c[i]].size(); j++) {
      if (j > 0) ret = ret * solve(v[c[i]][j - 1] + 1, v[c[i]][j] - 1) % MOD;
    }
    a[k++] = c[i];
    i = v[c[i]].back() + 1;
  }
  for (int i = 0; i < k; i++)
    for (int j = 0; j < k; j++) dp[i][j] = -1;
  ret = ret * calc(0, k - 1) % MOD;
  return ret;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; i++) {
    scanf("%d", &c[i]);
    c[i]--;
    v[c[i]].push_back(i);
  }
  set<int> s;
  s.insert(-1);
  s.insert(m);
  long long res = 1;
  for (int i = 0; i < n; i++) {
    int L = v[i].front(), R = v[i].back();
    int d = *s.lower_bound(L);
    if (d <= R) {
      printf("0\n");
      return 0;
    }
    for (int x : v[i]) s.insert(x);
  }
  printf("%I64d\n", solve(0, m - 1));
  return 0;
}