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 read() {
  int r = 0, t = 1, c = getchar();
  while (c < '0' || c > '9') {
    t = c == '-' ? -1 : 1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    r = r * 10 + c - 48;
    c = getchar();
  }
  return r * t;
}
const int N = 510, M = 1000010, mod = 998244353;
long long n, m, f[N][N], a[M], nex[M], lst[N], b[N], k, vis[N][N];
long long dfs2(long long l, long long r) {
  if (l > r) return 1;
  if (vis[l][r]) return f[l][r];
  vis[l][r] = 1;
  long long& ans = f[l][r];
  ans = 0;
  long long p = l;
  for (int i = (l + 1), end_i = (r); i <= end_i; i++)
    if (b[i] < b[p]) p = i;
  long long ans1 = 0, ans2 = 0;
  for (int i = (l), end_i = (p); i <= end_i; i++)
    ans1 = (ans1 + dfs2(l, i - 1) * dfs2(i, p - 1)) % mod;
  for (int i = (p), end_i = (r); i <= end_i; i++)
    ans2 = (ans2 + dfs2(p + 1, i) * dfs2(i + 1, r)) % mod;
  return ans = ans1 * ans2 % mod;
}
long long dfs1(long long l, long long r, long long mx) {
  if (l > r) return 1;
  int p = l;
  long long ans = 1;
  while (p <= r) {
    if (a[p] < mx) return 0;
    while (nex[p]) {
      if (nex[p] > r) return 0;
      ans = ans * dfs1(p + 1, nex[p] - 1, a[p]) % mod;
      p = nex[p];
    }
    p++;
  }
  p = l;
  k = 0;
  while (p <= r) {
    b[++k] = a[p];
    p = lst[a[p]] + 1;
  }
  for (int i = (1), end_i = (k); i <= end_i; i++)
    for (int j = (1), end_j = (k); j <= end_j; j++) vis[i][j] = 0;
  return ans * dfs2(1, k);
}
int main() {
  n = read();
  m = read();
  for (int i = (1), end_i = (m); i <= end_i; i++) {
    a[i] = read();
    nex[lst[a[i]]] = i;
    lst[a[i]] = i;
  }
  printf("%d\n", (int)dfs1(1, m, 0));
  return 0;
}