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 int maxn = 1111, mod = 998244353;
inline long long read() {
  char ch = getchar();
  long long x = 0, f = 0;
  while (ch < '0' || ch > '9') f |= ch == '-', ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return f ? -x : x;
}
int n, m, m_, a[maxn], lft[maxn], rig[maxn], f[maxn][maxn];
int main() {
  n = read();
  m_ = read();
  while (m_--) {
    int x = read();
    if (x != a[m]) a[++m] = x;
  }
  if (m > 2 * n) return puts("0"), 0;
  memset(lft, 0x3f, sizeof(lft));
  for (int i = (1); i <= (m); i++)
    lft[a[i]] = min(lft[a[i]], i), rig[a[i]] = max(rig[a[i]], i);
  for (int i = (1); i <= (m); i++)
    if (lft[a[i]] == i && rig[a[i]] == i) f[i][i] = 1;
  for (int i = (0); i <= (m); i++) f[i + 1][i] = f[i + 2][i] = 1;
  for (int l = (2); l <= (m); l++)
    for (int i = (1); i <= (m - l + 1); i++) {
      int j = i + l - 1, mn = i;
      for (int k = (i + 1); k <= (j); k++)
        if (a[k] < a[mn]) mn = k;
      int x = lft[a[mn]], y = rig[a[mn]], s1 = 0, s2 = 0, pre = 0;
      if (x < i || y > j) continue;
      for (int k = (i); k <= (x); k++)
        s1 = (s1 + 1ll * f[i][k - 1] * f[k][x - 1]) % mod;
      for (int k = (y); k <= (j); k++)
        s2 = (s2 + 1ll * f[y + 1][k] * f[k + 1][j]) % mod;
      f[i][j] = 1ll * s1 * s2 % mod;
      for (int k = (i); k <= (j); k++)
        if (a[k] == a[mn]) {
          if (pre) f[i][j] = 1ll * f[i][j] * f[pre + 1][k - 1] % mod;
          pre = k;
        }
    }
  printf("%d\n", f[1][m]);
}