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 INF = 1e9 + 10;
const int MOD = 998244353;
void add(int &a, int b) {
  a += b;
  if (a >= MOD) {
    a -= MOD;
  }
}
int sum(int a, int b) { return (a + b) % MOD; }
int mult(int a, int b) { return (a * 1ll * b) % MOD; }
const int N = 1010;
vector<int> have[N];
int dp[N][N], s1[N][N], s2[N][N], L[N], R[N], arr[N], n;
signed main() {
  ios_base::sync_with_stdio(0);
  int m;
  cin >> n >> m;
  int sz = 0;
  for (int i = 1; i <= m; i++) {
    int x;
    cin >> x;
    if (!sz || arr[sz] != x) {
      arr[++sz] = x;
      if (sz > 1000) {
        cout << 0 << '\n';
        exit(0);
      }
    }
  }
  for (int i = 1; i <= sz; i++) {
    if (!L[arr[i]]) {
      L[arr[i]] = i;
    }
    R[arr[i]] = i;
    have[arr[i]].push_back(i);
  }
  for (int l = 0; l <= sz + 1; l++) {
    for (int r = 0; r <= sz + 1; r++) {
      dp[l][r] = 1;
    }
  }
  for (int color = n; color >= 1; color--) {
    for (int l = 1; l <= L[color]; l++) {
      for (int l2 = l; l2 <= L[color]; l2++) {
        add(s1[color][l], mult(dp[l2][L[color] - 1], dp[l][l2 - 1]));
      }
    }
    for (int r = R[color]; r <= sz; r++) {
      for (int r2 = R[color]; r2 <= r; r2++) {
        add(s2[color][r], mult(dp[R[color] + 1][r2], dp[r2 + 1][r]));
      }
    }
    int sm = 1;
    for (int i = 0; i + 1 < (int)(have[color]).size(); i++) {
      sm = mult(sm, dp[have[color][i] + 1][have[color][i + 1] - 1]);
    }
    for (int l = 1; l <= sz; l++) {
      for (int r = l; r <= sz; r++) {
        if (l > R[color] || r < L[color]) {
          continue;
        } else if (max(l, L[color]) != L[color] ||
                   min(r, R[color]) != R[color]) {
          dp[l][r] = 0;
        } else {
          dp[l][r] = mult(sm, mult(s1[color][l], s2[color][r]));
        }
      }
    }
  }
  cout << dp[1][sz] << '\n';
}