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 mod = 998244353;
int n, m, cnt;
const int maxn = 505, maxm = maxn * 2;
int a[maxm];
vector<int> add[maxn];
long long dp[maxm][maxm];
int st[maxm][20];
void init() {
  for (int i = 1; i <= cnt; i++) {
    st[i][0] = a[i];
  }
  for (int i = 1; (1 << i) <= cnt; i++) {
    for (int j = 1; j + (1 << i) - 1 <= cnt; j++) {
      st[j][i] = min(st[j][i - 1], st[j + (1 << (i - 1))][i - 1]);
    }
  }
}
int mi(int l, int r) {
  int kk = (int)(log((double)(r - l + 1)) / log(2.0));
  return min(st[l][kk], st[r - (1 << kk) + 1][kk]);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cin >> n >> m;
  cnt = 0;
  int v;
  for (int i = 1; i <= m; i++) {
    cin >> v;
    if (v == a[cnt]) continue;
    if (cnt > 2 * n) {
      cout << 0 << endl;
      return 0;
    }
    a[++cnt] = v;
    add[v].push_back(cnt);
  }
  a[0] = a[cnt + 1] = 1e9;
  init();
  for (int i = 0; i <= cnt; i++) {
    dp[i + 1][i] = 1;
  }
  for (int l = 1; l <= cnt; l++) {
    for (int a = 1; a + l - 1 <= cnt; a++) {
      int x = mi(a, a + l - 1);
      int j = add[x][0];
      int k = add[x].back();
      if (j < a || k >= a + l) {
        dp[a][a + l - 1] = 0;
        continue;
      }
      long long lh = 0, rh = 0;
      for (int u = a; u <= j; u++) {
        lh += dp[a][u - 1] * dp[u][j - 1];
        lh %= mod;
      }
      for (int v = k + 1; v <= a + l; v++) {
        rh += dp[k + 1][v - 1] * dp[v][a + l - 1];
        rh %= mod;
      }
      dp[a][a + l - 1] = lh * rh % mod;
      for (unsigned int m = 0; m < add[x].size() - 1; m++) {
        dp[a][a + l - 1] *= dp[add[x][m] + 1][add[x][m + 1] - 1];
        dp[a][a + l - 1] %= mod;
      }
    }
  }
  cout << dp[1][cnt] % mod << endl;
  return 0;
}