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;
vector<int> st[10005];
long long dp[1005][1005];
int Minid[1005][1005];
int Minnum[1005][1005];
int a[1000005], tmp[1000005];
int mo = 998244353;
int L[1005], R[1005];
int main() {
  int n, m, ln = 0;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) scanf("%d", &tmp[i]);
  for (int i = 1; i <= m; i++) {
    if (tmp[i] == tmp[i - 1]) continue;
    a[++ln] = tmp[i];
  }
  if (ln > 2 * n) return !printf("0");
  for (int i = 1; i <= ln; i++) {
    if (L[a[i]] == 0) L[a[i]] = i;
    R[a[i]] = i;
    st[a[i]].push_back(i);
  }
  memset(Minnum, 0x3f, sizeof(Minnum));
  for (int i = 1; i <= ln; i++)
    for (int j = i; j <= ln; j++)
      for (int k = i; k <= j; k++) {
        if (Minnum[i][j] > a[k]) {
          Minnum[i][j] = a[k];
        }
      }
  for (int i = 0; i <= ln + 2; i++)
    for (int j = 0; j <= ln + 2; j++) dp[i][j] = 1;
  for (int len = 1; len <= ln; len++) {
    for (int i = 1; i <= ln - len + 1; i++) {
      int j = i + len - 1;
      bool flg = true;
      int Min = Minnum[i][j];
      if (L[Min] < i || R[Min] > j) {
        dp[i][j] = 0;
        continue;
      }
      long long Lans = 0, Rans = 0;
      for (int l = i; l <= L[Min]; l++) {
        Lans += dp[i][l - 1] * dp[l][L[Min] - 1] % mo;
        Lans %= mo;
      }
      for (int r = R[Min]; r <= j; r++) {
        Rans += dp[R[Min] + 1][r] * dp[r + 1][j];
        Rans %= mo;
      }
      dp[i][j] = Lans * Rans % mo;
      for (int k = 0; k < st[Min].size() - 1; k++) {
        dp[i][j] *= dp[st[Min][k] + 1][st[Min][k + 1] - 1];
        dp[i][j] %= mo;
      }
    }
  }
  printf("%lld\n", dp[1][ln]);
}