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;
int n, m;
int b[1222333];
int p1[555], p2[555];
int c[1111];
long long z = 1;
int vv[1111][1111];
long long dd[1111][1111];
long long calc(int a, int b) {
  if (a > b) return 1;
  if (vv[a][b]) return dd[a][b];
  int pi = a;
  for (int i = a; i <= b; i++) {
    if (c[i] < c[pi]) pi = i;
  }
  int g = c[pi];
  if (p1[g] < a || p2[g] > b) {
    vv[a][b] = 1;
    dd[a][b] = 0;
    return 0;
  }
  if (a == b) {
    vv[a][b] = 1;
    dd[a][b] = 1;
    return 1;
  }
  int pp = p1[g];
  long long du = 1;
  int ua = p1[g];
  int ub = p2[g];
  for (int i = ua + 1; i <= ub; i++) {
    if (c[i] == g) {
      du *= calc(pp + 1, i - 1);
      du %= 998244353;
      pp = i;
    }
  }
  pi = p1[g];
  long long d1 = 0;
  for (int i = a - 1; i < pi; i++) {
    d1 += calc(a, i) * calc(i + 1, pi - 1);
    d1 %= 998244353;
  }
  pi = p2[g];
  long long d2 = 0;
  for (int i = pi; i <= b; i++) {
    d2 += calc(pi + 1, i) * calc(i + 1, b);
    d2 %= 998244353;
  }
  vv[a][b] = 1;
  dd[a][b] = du * d1 % 998244353 * d2 % 998244353;
  return dd[a][b];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    cin >> b[i];
    if (!p1[b[i]]) p1[b[i]] = i;
    p2[b[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (p2[i] < p1[j]) continue;
      if (p2[j] < p1[i]) continue;
      if (p1[i] < p1[j] && p2[j] < p2[i]) continue;
      cout << 0 << "\n";
      return 0;
    }
  }
  int k = 0;
  for (int i = 1; i <= m; i++) {
    if (i == 1 || b[i - 1] != b[i]) {
      k++;
      c[k] = b[i];
    }
  }
  for (int i = 1; i <= n; i++) p1[i] = p2[i] = 0;
  for (int i = 1; i <= k; i++) {
    if (!p1[c[i]]) p1[c[i]] = i;
    p2[c[i]] = i;
  }
  cout << calc(1, k) << "\n";
}