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 long long MOD = 998244353;
long long add(long long a, long long b) { return (a + b) % MOD; }
long long mul(long long a, long long b) { return (a * b) % MOD; }
int n, m, c[501];
int lft[501], rht[501];
long long dp[501][501];
bool within(int x, int l, int r) { return x >= l && x <= r; }
long long eva(int l, int r) {
  if (l >= r) {
    return 1;
  }
  if (dp[l][r] != -1) return dp[l][r];
  int low = 501;
  for (int i = 1; i <= n; i++) {
    if ((within(lft[i], l, r) != within(rht[i], l, r))) {
      dp[l][r] = 0;
      return dp[l][r];
    }
    if (within(lft[i], l, r)) low = min(low, i);
  }
  long long res_1 = 0, res_2 = 0;
  for (int i = l; i <= lft[low]; i++) {
    long long res1 = eva(l, i - 1);
    long long res2 = eva(i, lft[low] - 1);
    res_1 = add(res_1, mul(res1, res2));
  }
  for (int j = rht[low]; j <= r; j++) {
    long long res3 = eva(rht[low] + 1, j);
    long long res4 = eva(j + 1, r);
    res_2 = add(res_2, mul(res3, res4));
  }
  dp[l][r] = mul(res_1, res_2);
  return dp[l][r];
}
int main(void) {
  cin >> n >> m;
  memset(lft, -1, sizeof(rht));
  memset(rht, -1, sizeof(rht));
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i < n; i++) {
    cin >> c[i];
    if (lft[c[i]] == -1) lft[c[i]] = i;
    rht[c[i]] = max(rht[c[i]], i);
  }
  cout << eva(0, n - 1) << endl;
}