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 N = 1e3 + 20;
const int oo = 1e9;
const int MOD = 998244353;
inline int mul(int a, int b) { return 1ll * a * b % MOD; }
inline int add(int a, int b) { return a + b < MOD ? a + b : a + b - MOD; }
int dp[N][N];
int a[N], first[N], lst[N];
int go(int l, int r) {
  if (l > r) return 1;
  int &R = dp[l][r];
  if (R == -1) {
    R = 0;
    int mn = *min_element(a + l, a + r + 1);
    if (first[mn] < l || lst[mn] > r)
      R = 0;
    else {
      int left = 0, right = 0, mid = 1;
      for (int i = l; i <= first[mn]; i++) {
        left = add(left, mul(go(l, i - 1), go(i, first[mn] - 1)));
      }
      int last = first[mn];
      for (int i = first[mn]; i <= lst[mn]; i++) {
        if (a[i] == mn) {
          mid = mul(mid, go(last + 1, i - 1));
          last = i;
        }
      }
      for (int i = lst[mn]; i <= r; i++) {
        right = add(right, mul(go(lst[mn] + 1, i), go(i + 1, r)));
      }
      R = mul(left, mul(mid, right));
    }
  }
  return R;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  while (cin >> n >> m) {
    for (int i = 0; i < m; i++) cin >> a[i];
    m = unique(a, a + m) - a;
    if (m > 2 * n) {
      cout << 0 << '\n';
    } else {
      memset(first, -1, sizeof first);
      for (int i = 0; i < m; i++) {
        if (first[a[i]] == -1) first[a[i]] = i;
        lst[a[i]] = i;
      }
      memset(dp, -1, sizeof dp);
      cout << go(0, m - 1) << '\n';
    }
  }
}