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 int64_t MOD = 998244353;
void add(int64_t& a, int64_t b) { a = (a + b) % MOD; }
void mul(int64_t& a, int64_t b) { a = a * b % MOD; }
int main() {
  int N, M, A[500], pos[500];
  cin >> N >> M;
  for (int i = 0; i < N; i++) {
    cin >> A[i];
    A[i]--;
    pos[A[i]] = i;
  }
  static int64_t dp[501][501];
  dp[pos[N - 1]][pos[N - 1]] = 1;
  for (int i = N - 2; i >= 0; i--) {
    int p = pos[i];
    vector<int64_t> L(N), R(N);
    int lt = p, rt = p;
    while (lt > 0 && A[lt - 1] > i) lt--;
    while (rt < N - 1 && A[rt + 1] > i) rt++;
    L[p] = R[p] = 1;
    for (int j = p; j >= lt; j--) {
      for (int k = lt; k < j; k++) add(L[k], dp[k][j - 1] * L[j]);
    }
    for (int j = p; j <= rt; j++) {
      for (int k = rt; k > j; k--) add(R[k], dp[j + 1][k] * R[j]);
    }
    for (int l = lt; l <= p; l++)
      for (int r = p; r <= rt; r++) add(dp[l][r], L[l] * R[r]);
  }
  static int64_t DP[501];
  DP[0] = 1;
  for (int i = 0; i < N; i++)
    for (int j = i; j < N; j++) add(DP[j + 1], DP[i] * dp[i][j]);
  cout << DP[N] << endl;
  return 0;
}