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 M = 1e6 + 9, N = 1003, mod = 998244353;
int n, m, x, p, mn, last, c[M], t[N], cnt[N], dp[N][N];
int calc(int l, int r) {
  int mn_el = N, mn_an = N, mx_an = 0, last_an = -1;
  int res = 1, ans1 = 0, ans2 = 0;
  if (l >= r) return 1;
  if (dp[l][r] != -1) return dp[l][r];
  for (int i = l; i < r + 1; i++) {
    mn_el = min(mn_el, c[i]);
    ;
    cnt[c[i]] = 0;
  }
  for (int i = l; i < r + 1; i++) cnt[c[i]]++;
  for (int i = l; i < r + 1; i++)
    if (cnt[c[i]] != t[c[i]]) return dp[l][r] = 0;
  for (int i = l; i < r + 1; i++)
    if (c[i] == mn_el) {
      if (last_an != -1) res = 1ll * res * calc(last_an + 1, i - 1) % mod;
      last_an = i;
      mx_an = max(mx_an, i);
      ;
      mn_an = min(mn_an, i);
      ;
    }
  for (int i = l - 1; i < mn_an; i++)
    ans1 = (ans1 + 1ll * calc(l, i) * calc(i + 1, mn_an - 1) % mod) % mod;
  for (int i = mx_an; i < r + 1; i++)
    ans2 = (ans2 + 1ll * calc(mx_an + 1, i) * calc(i + 1, r) % mod) % mod;
  dp[l][r] = 1ll * res * ans1 % mod * ans2 % mod;
  return dp[l][r];
}
int main() {
  for (int i = 0; i < N; i++)
    for (int j = 0; j < N; j++) dp[i][j] = -1;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    scanf("%d", &x);
    ;
    if (x != last) c[p++] = x;
    last = x;
  }
  m = p;
  if (n * 2 <= m) return cout << 0, 0;
  for (int i = 0; i < m; i++) {
    t[c[i]]++;
    mn = N;
    for (int j = i + 2; j < m; j++) {
      mn = min(mn, c[j - 1]);
      ;
      if (c[i] == c[j] && mn < c[i]) return cout << 0, 0;
    }
  }
  cout << calc(0, m - 1);
}