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 MOD = 998244353;
const int MAXM = 1000005;
const int SIZE = 1005;
int orig[MAXM];
int arr[MAXM];
int memo[SIZE][SIZE];
bool been[SIZE][SIZE];
int occ1[SIZE], occ2[SIZE];
int add(int a, int b) { return (a + b) % MOD; }
int mul(int a, int b) { return (long long)a * b % MOD; }
int solve(int l, int r) {
  if (l > r) return 1;
  if (been[l][r]) return memo[l][r];
  been[l][r] = true;
  int mn = *min_element(arr + l, arr + r + 1);
  int focc = occ1[mn], locc = occ2[mn];
  if (focc < l || locc > r) return memo[l][r] = 0;
  int ml = 1;
  vector<int> occs;
  for (int i = l; i <= r; i++) {
    if (arr[i] == mn) occs.push_back(i);
  }
  for (int i = 1; i < (int)occs.size(); i++) {
    ml = mul(ml, solve(occs[i - 1] + 1, occs[i] - 1));
  }
  int sum_l = 0;
  for (int ll = l; ll <= occs[0]; ll += 1) {
    sum_l = add(sum_l, mul(solve(l, ll - 1), solve(ll, occs[0] - 1)));
  }
  int sum_r = 0;
  for (int rr = occs.back(); rr <= r; rr += 1) {
    sum_r = add(sum_r, mul(solve(occs.back() + 1, rr), solve(rr + 1, r)));
  }
  int ans = mul(ml, mul(sum_r, sum_l));
  return memo[l][r] = ans;
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d", &orig[i]);
  }
  int real_m = 0;
  for (int i = 1; i <= n; i++) {
    if (orig[i] != orig[i - 1]) {
      arr[++real_m] = orig[i];
    }
  }
  for (int i = 1; i <= real_m; i++) {
    occ2[arr[i]] = i;
    if (occ1[arr[i]] == 0) occ1[arr[i]] = i;
  }
  if (real_m > 2 * n) {
    return 0 * puts("0");
  }
  printf("%d\n", solve(1, n));
  return 0;
}