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 _r() {
  int o = 0, f = 0;
  char t = getchar();
  while (t < 48 || t > 57) f |= (t == '-'), t = getchar();
  while (t > 47 && t < 58) o = o * 10 + t - 48, t = getchar();
  return f ? -o : o;
}
void setIO() {
  ios::sync_with_stdio(false);
  cin.tie(0);
}
int n, m, A[1000005], M[1005][1005];
long long F[1005][1005], mod = 998244353;
long long f(int i, int j) { return i > j ? 1 : F[i][j]; }
vector<int> P[1000005];
int L[1000005], R[1000005];
int main() {
  setIO();
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    cin >> A[i];
    if (A[i] == A[i - 1]) i--, m--;
  }
  if (m > 2 * n) return cout << 0, 0;
  for (int i = 1; i <= m; i++) {
    P[A[i]].push_back(i);
    if (!L[A[i]])
      L[A[i]] = i;
    else
      L[A[i]] = min(L[A[i]], i);
    if (!R[A[i]])
      R[A[i]] = i;
    else
      R[A[i]] = max(R[A[i]], i);
  }
  for (int i = 1; i <= m; i++) {
    M[i][i] = A[i];
    for (int j = i + 1; j <= m; j++) M[i][j] = min(M[i][j - 1], A[j]);
  }
  for (int j = 1; j <= m; j++) {
    for (int i = j; i >= 1; i--) {
      int g = 0;
      for (int k = i; k <= j; k++) {
        int c = A[k];
        if (L[c] < i || R[c] > j) g = 1;
      }
      if (g) continue;
      int c = M[i][j];
      long long l = 0, r = 0, o = 1;
      for (int k = i - 1; k < L[c]; k++)
        l += f(i, k) * f(k + 1, L[c] - 1), l %= mod;
      for (int k = R[c]; k <= j; k++)
        r += f(R[c] + 1, k) * f(k + 1, j), r %= mod;
      if (P[c].size() > 1)
        for (int i = 0; i < P[c].size() - 1; i++) {
          o = o * f(P[c][i] + 1, P[c][i + 1] - 1);
          o %= mod;
        }
      F[i][j] = l * r % mod;
      F[i][j] = F[i][j] * o % mod;
    }
  }
  cout << F[1][m];
}