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 = 505;
const int maxm = 1000005;
const int mod = 998244353;
int n, m, c[maxm], l[maxn], r[maxn];
long long f[maxn << 2][maxn << 2];
long long dp(int L, int R) {
  if (L == R + 1) return 1;
  if (L > R) return 0;
  if (~f[L][R]) return f[L][R];
  if (L == R) return f[L][R] = (l[c[L]] == L && r[c[R]] == R);
  int pos = L;
  for (int i = L + 1; i <= R; i++)
    if (c[i] < c[pos]) pos = i;
  if (l[c[pos]] < L || R < r[c[pos]]) return f[L][R] = 0;
  long long a = 0, b = 0;
  for (int i = L; i <= l[c[pos]]; i++)
    a = (a + dp(L, i - 1) * dp(i, l[c[pos]] - 1)) % mod;
  for (int i = R; i >= r[c[pos]]; i--)
    b = (b + dp(i + 1, R) * dp(r[c[pos]] + 1, i)) % mod;
  f[L][R] = a * b % mod;
  for (int i = pos, j; i != r[c[pos]];) {
    j = i, i++;
    while (c[i] != c[pos]) i++;
    f[L][R] = f[L][R] * dp(j + 1, i - 1) % mod;
  }
  return f[L][R];
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) scanf("%d", &c[i]);
  m = unique(c + 1, c + m + 1) - c - 1;
  for (int i = 1; i <= n; i++) l[i] = m + 1, r[i] = 0;
  for (int i = 1; i <= m; i++)
    l[c[i]] = min(l[c[i]], i), r[c[i]] = max(r[c[i]], i);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j < i; j++)
      if (!(r[i] < l[j] || (l[j] < l[i] && r[i] < r[j]) || r[j] < l[i])) {
        printf("0\n");
        return 0;
      }
  memset(f, -1, sizeof(f));
  printf("%lld\n", dp(1, m));
  return 0;
}