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 inf = 0x3f3f3f3f, mod = 998244353;
const double pi = 3.1415926535897932, eps = 1e-6;
int n, m, f[505][505], pos[505][505], a[505];
int mul(int x, int y) { return (long long)x * y % mod; }
int dfs(int l, int r) {
  if (l >= r) return 1;
  int &res = f[l][r];
  if (res != -1) return res;
  res = 0;
  int p = pos[l][r];
  long long sum1 = 0, sum2 = 0;
  for (int(i) = (l); (i) <= (p); (i)++) {
    sum1 += mul(dfs(l, i - 1), dfs(i, p - 1));
    if (sum1 >= mod) sum1 -= mod;
  }
  for (int(i) = (p); (i) <= (r); (i)++) {
    sum2 += mul(dfs(p + 1, i), dfs(i + 1, r));
    if (sum2 >= mod) sum2 -= mod;
  }
  res = mul(sum1, sum2);
  return res;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int(i) = (1); (i) <= (m); (i)++) scanf("%d", a + i);
  for (int(i) = (1); (i) <= (m); (i)++) {
    pos[i][i] = i;
    for (int(j) = (i + 1); (j) <= (m); (j)++)
      if (a[j] < a[pos[i][j - 1]])
        pos[i][j] = j;
      else
        pos[i][j] = pos[i][j - 1];
  }
  for (int(i) = (1); (i) <= (m); (i)++)
    for (int(j) = (1); (j) <= (m); (j)++) f[i][j] = -1;
  printf("%d\n", dfs(1, m));
  return 0;
}