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;
inline int read() {
  int X = 0, w = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') w = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') X = X * 10 + c - '0', c = getchar();
  return X * w;
}
const int N = 500 + 10;
const int mod = 998244353;
inline void add(int& x, int y) { x = (x + y) % mod; }
int n, m, c[N];
int dp[N][N];
int main() {
  n = read(), m = read();
  for (register int i = 1; i <= m; ++i) c[i] = read();
  for (register int i = 0; i <= m; ++i) dp[i + 1][i] = 1;
  for (register int l = 1; l <= m; ++l)
    for (register int i = 1, j = i + l - 1; j <= m; ++i, ++j) {
      int mn = 1e9, p;
      for (register int k = i; k <= j; ++k)
        if (c[k] < mn) mn = c[k], p = k;
      int L = 0, R = 0;
      for (register int k = i; k <= p; ++k)
        add(L, 1ll * dp[i][k - 1] * dp[k][p - 1] % mod);
      for (register int k = p; k <= j; ++k)
        add(R, 1ll * dp[p + 1][k] * dp[k + 1][j] % mod);
      add(dp[i][j], 1ll * L * R % mod);
    }
  printf("%d\n", dp[1][m]);
  return 0;
}