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 dp[1024][1024], mm[1024][1024], in[1024], out[1024];
int mod = 998244353;
int go(int l, int r, vector<int>& v) {
  int p = -1;
  long long ret;
  if (l == r + 1) return 1;
  if (dp[l][r] != -1) return dp[l][r];
  dp[l][r] = 0;
  int mo = l;
  for (int i = l; i <= r; i++) {
    int xpos = mm[l][i];
    int x = v[xpos];
    long long lret = ret;
    mo = max(mo, out[v[i]]);
    if (p < 0 || v[p] != x) {
      p = xpos;
      ret = go(l, xpos - 1, v);
      lret = ret;
      if (i >= mo) {
        lret *= go(xpos + 1, i, v);
        lret %= mod;
      }
    } else if (v[i] == x) {
      ret *= go(p + 1, i - 1, v);
      ret %= mod;
      p = i;
      lret = ret;
    } else {
      lret = ret;
      if (i >= mo) {
        lret *= go(p + 1, i, v);
        lret %= mod;
      }
    }
    if (i < mo) continue;
    lret *= go(i + 1, r, v);
    lret %= mod;
    dp[l][r] += lret;
    dp[l][r] %= mod;
  }
  return dp[l][r];
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  vector<int> v(m);
  for (auto& e : v) scanf("%d", &e);
  int r = 0;
  for (int i = 0; i < m;) {
    int j;
    for (j = i; j < m && v[j] == v[i]; j++)
      ;
    v[r++] = v[i];
    i = j;
  }
  m = r;
  for (int i = 0; i < (int)m; i++) {
    memset(dp[i], -1, sizeof(dp[i]));
  }
  for (int i = 0; i < (int)n + 1; i++) {
    in[i] = m, out[i] = 0;
  }
  for (int i = 0; i < (int)m; i++) {
    in[v[i]] = min(in[v[i]], i);
    out[v[i]] = max(out[v[i]], i);
  }
  for (int i = 0; i < m; i++) {
    mm[i][i] = i;
    for (int j = i + 1; j < m; j++) {
      mm[i][j] = mm[i][j - 1];
      if (v[j] < v[mm[i][j]]) mm[i][j] = j;
    }
  }
  for (int i = 0; i < (int)m; i++)
    if (v[mm[in[v[i]]][out[v[i]]]] < v[i]) {
      printf("0\n");
      return 0;
    }
  printf("%d\n", go(0, m - 1, v));
  return 0;
}