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;
int a[1000005], b[1000005], v[505];
int f[505 << 1][505 << 1];
set<int> s[505];
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  return x * f;
}
bool vis[505];
int dp(int l, int r) {
  if (l >= r) return 1;
  if (~f[l][r]) return f[l][r];
  int &ret = f[l][r], res = 1, mn = n + 1;
  ret = 0;
  for (int i = l; i <= r; i++) mn = min(mn, a[i]);
  for (set<int>::iterator pre = s[mn].begin(), it = next(pre);
       it != s[mn].end(); pre = it++)
    res = 1ll * res * dp(*pre + 1, *it - 1) % mod;
  int sl = *s[mn].begin(), sr = *s[mn].rbegin();
  int sum[r - l + 1];
  int cnt = 0;
  for (int i = l; i <= r; i++) {
    if (!vis[a[i]]) vis[a[i]] = 1, v[++cnt] = a[i];
    sum[i - l] = 0;
  }
  for (int i = 1; i <= cnt; i++) {
    if (*s[v[i]].rbegin() < sl)
      sum[*s[v[i]].begin() + 1 - l]++, sum[*s[v[i]].rbegin() + 1 - l]--;
    if (*s[v[i]].begin() > sr)
      sum[*s[v[i]].begin() - l]++, sum[*s[v[i]].rbegin() - l]--;
  }
  for (int i = 1; i < r - l + 1; i++) sum[i] += sum[i - 1];
  for (int i = l; i <= r; i++) vis[a[i]] = 0;
  int res1 = 0, res2 = 0;
  for (int i = l; i <= sl; i++)
    if (!sum[i - l]) res1 = (res1 + 1ll * dp(l, i - 1) * dp(i, sl - 1)) % mod;
  for (int i = sr; i <= r; i++)
    if (!sum[i - l]) res2 = (res2 + 1ll * dp(sr + 1, i) * dp(i + 1, r)) % mod;
  ret = 1ll * res1 * res % mod * res2 % mod;
  return ret;
}
int main() {
  n = read(), m = read();
  for (int i = 1; i <= m; i++) a[i] = read();
  for (int i = 1; i <= m; i++)
    if (a[i] != a[i + 1]) b[++cnt] = a[i];
  m = cnt;
  for (int i = 1; i <= m; i++) a[i] = b[i];
  for (int i = 1; i <= m; i++) s[a[i]].insert(i);
  for (int i = 1; i <= n; i++) {
    int l = *s[i].begin(), r = *s[i].rbegin();
    for (int j = l; j <= r; j++)
      if (a[j] < i) {
        puts("0");
        return 0;
      }
  }
  memset(f, -1, sizeof(f));
  printf("%d\n", dp(1, m));
  return 0;
}