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, N = 510;
int n, m, a[N], st[N][12], g[N][12], dp[N][N];
int solve(int l, int r) {
  if (dp[l][r]) return dp[l][r];
  if (l > r) return 1;
  long long r1 = 0, r2 = 0;
  int pos, lo = log2(r - l + 1);
  if (st[l][lo] < st[r - (1 << lo) + 1][lo])
    pos = g[l][lo];
  else
    pos = g[r - (1 << lo) + 1][lo];
  for (int i = l; i <= pos; i++)
    r1 += 1ll * solve(l, i - 1) * solve(i, pos - 1) % mod;
  for (int i = pos; i <= r; i++)
    r2 += 1ll * solve(pos + 1, i) * solve(i + 1, r) % mod;
  r1 %= mod;
  r2 %= mod;
  return dp[l][r] = 1ll * r1 * r2 % mod;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), st[i][0] = a[i], g[i][0] = i;
  for (int i = 1; (1 << i) <= n; i++) {
    for (int j = 1; j + (1 << i) - 1 <= n; j++) {
      int y = j + (1 << (i - 1));
      if (st[j][i - 1] < st[y][i - 1])
        g[j][i] = g[j][i - 1];
      else
        g[j][i] = g[y][i - 1];
      st[j][i] = min(st[j][i - 1], st[y][i - 1]);
    }
  }
  printf("%d\n", solve(1, n));
  return 0;
}