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;
using i64 = long long;
const i64 P = 998244353;
i64 dp[1234][1234];
int a[1234567], n, m, t, posl[1234], posr[1234];
i64 solve(int l, int r) {
  if (dp[l][r] != -1) return dp[l][r];
  if (l > r) return 1;
  i64 &val = dp[l][r];
  val = 0;
  int minc = a[l];
  for (int j = l; j <= r; j++) {
    minc = min(minc, a[j]);
  }
  int prepos = -1;
  i64 tmpval = 1;
  for (int j = l; j <= r; j++) {
    if (a[j] == minc) {
      if (prepos != -1) {
        tmpval = tmpval * solve(prepos + 1, j - 1) % P;
      }
      prepos = j;
    }
  }
  i64 lval = 0;
  for (int j = l; j <= posl[minc]; j++) {
    lval = (lval + solve(l, j - 1) * solve(j, posl[minc] - 1)) % P;
  }
  i64 rval = 0;
  for (int j = posr[minc]; j <= r; j++) {
    rval = (rval + solve(posr[minc] + 1, j) * solve(j + 1, r)) % P;
  }
  return val = tmpval * lval % P * rval % P;
}
int main() {
  memset(dp, -1, sizeof(dp));
  scanf("%d%d", &n, &m);
  int c = 0;
  for (int i = 1; i <= m; i++) {
    int x;
    scanf("%d", &x);
    if (x != c) a[++t] = x;
    c = x;
  }
  if (t >= 2 * n) {
    printf("%d\n", 0);
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    int l = t + 1, r = 0;
    for (int j = 1; j <= t; j++) {
      if (a[j] == i) {
        l = min(l, j);
        r = max(r, j);
      }
    }
    for (int j = l; j <= r; j++)
      if (a[j] < i) {
        printf("%d\n", 0);
        return 0;
      }
    posl[i] = l;
    posr[i] = r;
  }
  printf("%lld\n", solve(1, t));
}