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 int64_t mod = 998244353;
int64_t mul(int64_t a, int64_t b) { return a * b % mod; }
int64_t add(int64_t a, int64_t b) { return (a + b) % mod; }
int64_t bpow(int64_t x, int64_t n) {
  return n ? n % 2 ? x * bpow(x, n - 1) % mod : bpow(x * x % mod, n / 2) : 1;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int64_t n, m;
  cin >> n >> m;
  int64_t c[m];
  vector<int64_t> rev[n];
  for (int64_t i = 0; i < m; i++) {
    cin >> c[i];
    rev[c[i] - 1].push_back(i);
  }
  int64_t dp[n + 1][n + 1][2];
  memset(dp, 0, sizeof(dp));
  for (int64_t i = 0; i < n; i++) {
    dp[i][i][0] = 1;
    dp[i][i][1] = 0;
  }
  for (int64_t ln = 1; ln < n; ln++) {
    for (int64_t i = 0; i + ln < n; i++) {
      int64_t j = i + ln;
      int64_t t = min_element(c + i, c + j + 1) - c;
      for (int64_t k = i; k < j; k++) {
        dp[i][j][1] =
            add(dp[i][j][1],
                mul(dp[i][k][0], add(dp[k + 1][j][1], dp[k + 1][j][0])));
      }
      if (t == i) {
        dp[i][j][0] = add(dp[i + 1][j][1], dp[i + 1][j][0]);
      } else if (t == j) {
        dp[i][j][0] = add(dp[i][j - 1][1], dp[i][j - 1][0]);
      } else {
        dp[i][j][0] = mul(add(dp[i][t - 1][1], dp[i][t - 1][0]),
                          add(dp[t + 1][j][1], dp[t + 1][j][0]));
      }
    }
  }
  cout << add(dp[0][n - 1][1], dp[0][n - 1][0]) << endl;
  return 0;
}