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;
long long tree[1501] = {INT_MAX};
vector<long long> target(501, INT_MAX);
long long dp[501][501];
map<long long, long long> clr;
void build(long long s, long long e, long long i) {
  if (s == e) {
    tree[i] = target[s];
    return;
  }
  long long mid = (s + e) / 2;
  build(s, mid, 2 * i);
  build(mid + 1, e, 2 * i + 1);
  tree[i] = min(tree[2 * i], tree[2 * i + 1]);
  return;
}
long long query(long long s, long long e, long long l, long long r,
                long long i) {
  if (l > e || r < s) return INT_MAX;
  if (l <= s && r >= e) {
    return tree[i];
  }
  long long mid = (s + e) / 2;
  return min(query(s, mid, l, r, 2 * i), query(mid + 1, e, l, r, 2 * i + 1));
}
long long solve(long long l, long long r, long long n) {
  if (l < 0 || r >= n) return 0;
  if (l == r) return 1;
  if (l > r) return 1;
  if (dp[l][r] != -1) return dp[l][r];
  long long cmin = query(0, n - 1, l, r, 1);
  long long ci = clr[cmin];
  long long ans = 0, left = 0, right = 0;
  for (long long a = l; a <= ci; a++) {
    left = (left + (solve(l, a - 1, n) * solve(a, ci - 1, n)) % 998244353) %
           998244353;
  }
  for (long long b = ci; b <= r; b++) {
    right = (right + (solve(ci + 1, b, n) * solve(b + 1, r, n)) % 998244353) %
            998244353;
  }
  dp[l][r] = (left * right) % 998244353;
  return dp[l][r];
}
void mymain() {
  long long n, m;
  cin >> n >> m;
  for (long long i = 0; i < n; i++) {
    cin >> target[i];
    clr[target[i]] = i;
  }
  build(0, n - 1, 1);
  for (long long i = 0; i < 501; i++)
    for (long long j = 0; j < 501; j++) dp[i][j] = -1;
  cout << solve(0, n - 1, n) << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long t;
  t = 1;
  for (long long tt = 0; tt < t; tt++) {
    mymain();
  }
  return 0;
}