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 n, m, tot, a[1000005];
long long dp[1005][1005];
vector<int> p[1005];
int rd() {
  int x = 0, tp = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') tp = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * tp;
}
long long getans(int l, int r) {
  if (l > r) return 1;
  if (dp[l][r] != -1) return dp[l][r];
  int mn = a[l], L, R, sz;
  for (int i = l + 1; i <= r; i++) mn = min(a[i], mn);
  long long nl = 0, nr = 0;
  sz = p[mn].size(), L = p[mn][0], R = p[mn][sz - 1];
  for (int i = l; i <= L; i++)
    (nl += getans(l, i - 1) * getans(i, L - 1) % 998244353) %= 998244353;
  for (int i = R; i <= r; i++)
    (nr += getans(R + 1, i) * getans(i + 1, r) % 998244353) %= 998244353;
  dp[l][r] = nl * nr % 998244353;
  for (int i = 0; i < sz - 1; i++)
    (dp[l][r] *= getans(p[mn][i] + 1, p[mn][i + 1] - 1)) %= 998244353;
  return dp[l][r];
}
int main() {
  memset(dp, -1, sizeof(dp));
  n = rd(), m = rd();
  for (int i = 1; i <= m; i++) a[i] = rd();
  tot = unique(a + 1, a + m + 1) - a - 1;
  for (int i = 1; i <= tot; i++) p[a[i]].push_back(i);
  if (tot > 2 * n) {
    printf("0\n");
    return 0;
  }
  printf("%lld\n", getans(1, tot));
  return 0;
}