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;
const long long N = 5e2 + 1;
const long long M = 1e6 + 1;
long long dp[2 * N][2 * N], min1[2 * N][2 * N], ar[2 * N], ar1[M];
bool check[2 * N][2 * N];
vector<long long> occur[N];
long long solve(long long l, long long r) {
  if (l > r) {
    return 1;
  }
  if (check[l][r]) {
    return dp[l][r];
  }
  long long y = min1[l][r];
  if (occur[y][0] < l || occur[y][occur[y].size() - 1] > r) {
    check[l][r] = true;
    return 0;
  }
  if (l == r) {
    check[l][r] = true;
    dp[l][r] = 1;
    return dp[l][r];
  }
  check[l][r] = true;
  long long tich = 1, sum = solve(l, occur[y][0] - 1),
            sum1 = solve(occur[y][occur[y].size() - 1] + 1, r), i;
  if (l <= occur[y][0] - 1) {
    sum *= 2;
  }
  if (occur[y][occur[y].size() - 1] + 1 <= r) {
    sum1 *= 2;
  }
  for (i = 0; i + 1 < occur[y].size(); i++) {
    tich *= solve(occur[y][i] + 1, occur[y][i + 1] - 1);
    tich %= mod;
  }
  for (i = l; i < occur[y][0] - 1; i++) {
    sum += solve(l, i) * solve(i + 1, occur[y][0] - 1);
    sum %= mod;
  }
  for (i = occur[y][occur[y].size() - 1] + 1; i < r; i++) {
    sum1 += solve(occur[y][occur[y].size() - 1] + 1, i) * solve(i + 1, r);
    sum1 %= mod;
  }
  dp[l][r] = ((sum * sum1) % mod * tich) % mod;
  return dp[l][r];
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, m, i, j, k, l = 0;
  cin >> n >> m;
  ar1[0] = -1;
  for (i = 1; i <= m; i++) {
    cin >> ar1[i];
    if (ar1[i] != ar1[i - 1]) {
      l++;
      ar[l] = ar1[i];
      if (l > 2 * n + 1) {
        cout << 0;
        exit(0);
      }
    }
  }
  n = l;
  for (i = 1; i <= n; i++) {
    min1[i][i] = ar[i];
    occur[ar[i]].push_back(i);
    for (j = i + 1; j <= n; j++) {
      min1[i][j] = min(min1[i][j - 1], ar[j]);
    }
  }
  cout << solve(1, n);
}