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;
string s;
const long long MOD = 998244353;
vector<int> a;
long long dp[501][501];
int n, m;
void calc(int i, int j) {
  if (i == j)
    dp[i][j] = 1;
  else {
    int mm = 1e9;
    int pos = -1;
    for (int k = i; k < j; k++) {
      if (mm > a[k]) {
        mm = a[k];
        pos = k;
      }
    }
    int k = pos;
    long long delta1[2];
    delta1[0] = 0;
    delta1[1] = 0;
    for (int x = i; x <= k; x++) {
      long long delta = 1;
      delta *= dp[i][x];
      delta %= MOD;
      delta *= dp[x][k];
      delta %= MOD;
      delta1[0] += delta;
      delta1[0] %= MOD;
    }
    for (int y = k + 1; y <= j; y++) {
      long long delta = 1;
      delta *= dp[k + 1][y];
      delta %= MOD;
      delta *= dp[y][j];
      delta %= MOD;
      delta1[1] += delta;
      delta1[1] %= MOD;
    }
    delta1[0] *= delta1[1];
    delta1[0] %= MOD;
    dp[i][j] += delta1[0];
    dp[i][j] %= MOD;
  }
}
void calc_dp() {
  for (int delta = 0; delta < n + 10; delta++) {
    for (int alp = 0; alp < n + 10; alp++) {
      if (alp + delta <= n) calc(alp, alp + delta);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  a.resize(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  calc_dp();
  cout << dp[0][n];
  return 0;
}