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 MAXM = 1e6 + 1, MAXN = 501, MOD = 998244353;
int n, m, c[MAXM], s[MAXN], d[MAXN], memo[MAXN][MAXN], last[MAXN], nxt[MAXM];
int dp(int l, int r) {
  if (!l) return 1;
  if (l != r) {
    if (d[l] >= d[r]) return 1;
    if (s[l] >= s[r]) return 1;
  }
  if (memo[l][r] != -1) {
    return memo[l][r];
  }
  int i = l;
  bool same = 1;
  for (int j = 1; j <= n; j++) {
    if (s[j] >= s[l] && s[j] <= d[r] && d[j] >= s[l] && d[j] <= d[r]) {
      if (j != l) same = 0;
      i = min(i, j);
    }
  }
  if (same) return memo[l][r] = 1;
  long long f = 0, k = 0;
  if (i == l)
    f = 1;
  else {
    int lo = c[s[i]];
    while (s[lo] >= s[l]) {
      if (!lo) break;
      long long tmp = 1;
      (tmp *= dp(lo, c[s[i] - 1])) %= MOD;
      (tmp *= dp(l, c[s[lo] - 1])) %= MOD;
      (f += tmp) %= MOD;
      lo = (s[lo] > 0 ? c[s[lo] - 1] : 0);
    }
  }
  if (i == r)
    k = 1;
  else {
    int hi = c[d[i]];
    while (s[hi] <= d[r]) {
      if (!hi) break;
      long long tmp = 1;
      (tmp *= dp(c[d[i] + 1], hi)) %= MOD;
      (tmp *= dp(c[d[hi] + 1], r)) %= MOD;
      (k += tmp) %= MOD;
      hi = (d[hi] < d[r] ? c[d[hi] + 1] : 0);
    }
  }
  long long res = (f * k) % MOD;
  int start = s[i];
  while (nxt[start] != -1) {
    if (nxt[start] - start > 1)
      (res *= dp(c[start + 1], c[nxt[start] - 1])) %= MOD;
    start = nxt[start];
  }
  return memo[l][r] = res;
}
bool wrong() {
  for (int i = 1; i <= n; i++) {
    for (int j = s[i]; j <= d[i]; j++) {
      if (c[j] < i) return 1;
    }
  }
  return 0;
}
void magic() {
  memset(d, 0, sizeof(d));
  for (int i = 1; i <= n; i++) s[i] = m;
  for (int i = 0; i < m; i++) {
    s[c[i]] = min(s[c[i]], i);
    d[c[i]] = max(d[c[i]], i);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  memset(last, -1, sizeof(last));
  memset(nxt, -1, sizeof(nxt));
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> c[i];
    if (last[c[i]] != -1) nxt[last[c[i]]] = i;
    last[c[i]] = i;
  }
  magic();
  if (wrong()) {
    cout << "0";
    exit(0);
  }
  memset(memo, -1, sizeof(memo));
  cout << dp(c[0], c[m - 1]);
  return 0;
}