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 N = 2e3 + 10;
const int mod = 998244353;
int f[N][N], g[N][N];
int mn[N][N], indl[N], indr[N];
bool bz[N][N];
int a[1000001];
int n, m;
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++) scanf("%d", &a[i]);
  int x = m;
  m = 1;
  for (int i = 2; i <= x; i++)
    if (a[i] != a[m]) a[++m] = a[i];
  for (int i = 1; i <= m; i++) {
    if (!indl[a[i]]) indl[a[i]] = i;
    indr[a[i]] = i;
  }
  if (m > 3 * n) {
    printf("0\n");
    return 0;
  }
  for (int i = 1; i <= n; i++)
    for (int j = indl[i]; j <= indr[i]; j++)
      if (a[j] < i) {
        printf("0\n");
        return 0;
      }
  for (int i = 1; i <= m; i++) {
    mn[i][i] = a[i];
    for (int j = i + 1; j <= m; j++) mn[i][j] = min(mn[i][j - 1], a[j]);
  }
  for (int i = 1; i <= m + 1; i++) f[i][i - 1] = g[i][i - 1] = 1;
  for (int i = 1; i <= m; i++)
    for (int l = 1; l + i - 1 <= m; l++) {
      int r = i + l - 1;
      int x = mn[l][r], idl = indl[x], idr = indr[x];
      if (idl > r || idr < l) continue;
      int v = 1;
      int lll = idl;
      while (lll != idr) {
        int rrr = lll + 1;
        while (a[rrr] != a[lll]) rrr++;
        v = (long long)v * f[lll + 1][rrr - 1] % mod;
        lll = rrr;
      }
      for (int mid = idr; mid <= r; mid++)
        g[l][r] = (g[l][r] + (long long)f[l][idl - 1] * f[mid + 1][r] % mod *
                                 v % mod * f[idr + 1][mid] % mod) %
                  mod;
      for (int ll = l; ll <= idl; ll++)
        f[l][r] = (f[l][r] + (long long)g[ll][r] * f[l][ll - 1]) % mod;
    }
  printf("%d\n", f[1][m]);
  return 0;
}