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 long long mod = 998244353;
const int MAX = 505;
int N, K, st[MAX][10];
int a[MAX], b[MAX];
long long pd[MAX][MAX];
void SparseTable() {
  for (int j = (0); j < (10); j++)
    for (int i = 1; (i + (1 << j) - 1) <= N; i++)
      st[i][j] = j ? min(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]) : a[i];
}
int query(int l, int r, int k) {
  return min(st[l][k], st[r - (1 << k) + 1][k]);
}
long long solve(int i, int j) {
  if (i > j) return 1;
  if (pd[i][j] != -1) return pd[i][j];
  int m = b[query(i, j, log2(j - i + 1))];
  long long left = 0, right = 0;
  for (int l = (i); l <= (m); l++)
    left = (left + solve(i, l - 1) * solve(l, m - 1)) % mod;
  for (int r = (m); r <= (j); r++)
    right = (right + solve(m + 1, r) * solve(r + 1, j)) % mod;
  return pd[i][j] = (left * right) % mod;
}
int main(int argc, char** argv) {
  scanf("%d%d", &N, &K);
  for (int i = (1); i <= (N); i++) scanf("%d", a + i);
  for (int i = (1); i <= (N); i++) b[a[i]] = i;
  SparseTable();
  memset(pd, -1, sizeof pd);
  printf("%I64d\n", solve(1, N));
  return 0;
}