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>
#pragma GCC optimize("O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int arr = 2e5 + 10;
const int ar = 2e3 + 10;
const long double pi = acos(-1);
const long double eps = 1e-10;
const long long md = 1e9 + 7;
int a[(int)(500 + 10)];
int dp[(int)(500 + 10)][(int)(500 + 10)];
int get_dp(int l, int r) {
  if (l > r) {
    return 1;
  }
  if (l == r) {
    return 1;
  }
  if (dp[l][r] != -1) {
    return dp[l][r];
  }
  dp[l][r] = 0;
  int pos = min_element(a + l, a + r + 1) - a;
  int sum1 = 0;
  int sum2 = 0;
  for (int k = l; k <= pos; k++) {
    sum1 +=
        1ll * get_dp(l, k - 1) * get_dp(k, pos - 1) % (long long)(998244353ll);
    if (sum1 >= (long long)(998244353ll)) {
      sum1 -= (long long)(998244353ll);
    }
  }
  for (int p = pos; p <= r; p++) {
    sum2 +=
        1ll * get_dp(pos + 1, p) * get_dp(p + 1, r) % (long long)(998244353ll);
    if (sum2 >= (long long)(998244353ll)) {
      sum2 -= (long long)(998244353ll);
    }
  }
  dp[l][r] = 1ll * sum1 * sum2 % (long long)(998244353ll);
  return dp[l][r];
}
int main() {
  for (int i = 0; i < (int)(500 + 10); i++) {
    for (int j = i; j < (int)(500 + 10); j++) {
      dp[i][j] = -1;
    }
  }
  int n;
  cin >> n >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  cout << get_dp(1, n) << "\n";
}