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;
int N, M;
int A[505], B[505][505];
long long dp[505][505];
long long f(int L, int R) {
  if (L > R) return 1;
  long long &ret = dp[L][R];
  if (ret != -1) return ret;
  ret = 0;
  int mid = B[L][R];
  long long LS = 0, RS = 0;
  for (int i = L; i <= mid; i++) {
    LS += f(L, i - 1) * f(i, mid - 1);
    LS %= 998244353;
  }
  for (int i = mid; i <= R; i++) {
    RS += f(mid + 1, i) * f(i + 1, R);
    RS %= 998244353;
  }
  ret = LS * RS % 998244353;
  return ret;
}
int main() {
  memset(dp, -1, sizeof(dp));
  scanf("%d %*d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", &A[i]);
  for (int i = 1; i <= N; i++) {
    int mn = 1234567890, idx = -1;
    for (int j = i; j <= N; j++) {
      if (mn > A[j]) mn = A[j], idx = j;
      B[i][j] = idx;
    }
  }
  printf("%lld", f(1, N));
  return 0;
}