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;
using ll = long long;
const int N = 1010, M = 1000006;
const ll mod = 998244353;
int n, m, c[M];
ll dp[N][N];
int cnt[N];
ll f(int L, int R) {
  if (L > R) return 1;
  ll &ans = dp[L][R];
  if (~ans) return ans;
  int id = L, id2 = L;
  for (int i = L + 1; i <= R; i++) {
    if (c[i] < c[id]) id = i;
    if (c[i] <= c[id2]) id2 = i;
  }
  int in_cnt = 0;
  for (int i = L; i <= R; i++)
    if (c[i] == c[id]) in_cnt++;
  if (in_cnt != cnt[c[id]]) return ans = 0;
  if (L == R) return ans = 1;
  ll a = 0, b = 0;
  for (int i = L; i <= id; i++) {
    a = (a + f(L, i - 1) * f(i, id - 1)) % mod;
  }
  for (int j = id2; j <= R; j++) {
    b = (b + f(id2 + 1, j) * f(j + 1, R)) % mod;
  }
  ll x = 1;
  for (int j = id; j < id2; j++) {
    int tmp = j + 1;
    while (tmp <= id2 and c[tmp] != c[id]) tmp++;
    x = x * f(j + 1, tmp - 1) % mod;
    j = tmp - 1;
  }
  return ans = a * b % mod * x % mod;
}
int main() {
  memset(dp, -1, sizeof dp);
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; i++) scanf("%d", c + i);
  {
    int j = 0;
    for (int i = 1; i < m; i++) {
      if (c[i] != c[j]) {
        c[++j] = c[i];
      }
    }
    m = j + 1;
  }
  if (m > 2 * n + 5) return printf("0\n"), 0;
  for (int i = 0; i < m; i++) cnt[c[i]]++;
  printf("%lld\n", f(0, m - 1));
}