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 dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};
const int maxn = 1005;
long long dp[maxn][maxn];
vector<int> v[maxn];
int a[1000005], book[maxn], lo[maxn][maxn], ma[maxn], mi[maxn];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) scanf("%d", &a[i]);
  int tot = 0;
  for (int i = 1; i <= m; i++) {
    if (i == 1 || a[i] != a[i - 1]) a[++tot] = a[i];
  }
  m = tot;
  if (m > 2 * n) return 0 * printf("0");
  for (int i = 1; i <= m; i++) {
    ma[a[i]] = max(i, ma[a[i]]);
    if (!mi[a[i]]) mi[a[i]] = i;
    v[a[i]].push_back(i);
  }
  for (int i = 1; i <= m; i++) {
    for (int j = 1; j <= m; j++) {
      if (i == j)
        lo[i][j] = a[i];
      else
        lo[i][j] = min(lo[i][j - 1], a[j]);
    }
  }
  for (int r = 1; r <= m; r++) {
    for (int l = r; l >= 1; l--) {
      int lll = mi[lo[l][r]], rr = ma[lo[l][r]];
      if (l == r) {
        dp[l][r] = !(lll < l || rr > r);
        dp[l][r - 1] = 1;
        dp[l + 1][r] = 1;
        continue;
      }
      if (lll < l || rr > r) {
        dp[l][r] = 0;
        continue;
      }
      long long left, right, mid;
      left = right = 0;
      mid = 1;
      int now = lo[l][r];
      for (int k = l; k <= lll; k++)
        left = (left + dp[l][k - 1] * dp[k][lll - 1]) % 998244353;
      for (int k = rr; k <= r; k++)
        right = (right + dp[k + 1][r] * dp[rr + 1][k]) % 998244353;
      for (int k = 1; k < v[now].size(); k++)
        mid = mid * dp[v[now][k - 1] + 1][v[now][k] - 1] % 998244353;
      dp[l][r] = 1ll * left * right % 998244353 * mid % 998244353;
    }
  }
  printf("%lld\n", dp[1][m]);
  return 0;
}