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 N = 1e3 + 3, M = 1e6 + 3, Mod = 998244353;
struct Mint {
  int res;
  Mint() { res = -1; }
  Mint(int a) : res(a) {}
  friend Mint operator+(Mint A, Mint B) {
    return (A.res + B.res >= Mod) ? (A.res + B.res - Mod) : (A.res + B.res);
  }
  friend Mint operator-(Mint A, Mint B) {
    B.res = Mod - B.res;
    return A + B;
  }
  friend Mint operator*(Mint A, Mint B) { return 1ll * A.res * B.res % Mod; }
};
int n, m, c[M], L[N], R[N], Mn[N][N];
Mint dp[N][N];
Mint F(int l, int r) {
  if (r <= l) return 1;
  if (~dp[l][r].res) return dp[l][r];
  if (L[Mn[l][r]] < l || R[Mn[l][r]] > r) return dp[l][r] = Mint(0);
  if (r - l <= 1) return dp[l][r] = Mint(1);
  Mint Lprd = Mint(0), Rprd = Mint(0), Mprd = Mint(1);
  for (register int i = (l); i <= (L[Mn[l][r]]); ++i)
    Lprd = Lprd + F(i, L[Mn[l][r]]) * F(l, i);
  for (register int i = (R[Mn[l][r]]); i <= (r); ++i)
    Rprd = Rprd + F(R[Mn[l][r]], i) * F(i, r);
  int lst = L[Mn[l][r]];
  for (register int i = (L[Mn[l][r]] + 1); i <= (R[Mn[l][r]]); ++i)
    if (c[i] == Mn[l][r]) Mprd = Mprd * F(lst + 1, i), lst = i;
  return dp[l][r] = Lprd * Rprd * Mprd;
}
int main() {
  scanf("%d%d", &n, &m);
  for (register int i = (1); i <= (m); ++i) scanf("%d", c + i);
  m = unique(c + 1, c + m + 1) - c - 1;
  if (m > 2 * n) return puts("0"), 0;
  for (register int i = (1); i <= (n); ++i) L[i] = m + 1, R[i] = 0;
  for (register int i = (1); i <= (m); ++i)
    L[c[i]] = min(L[c[i]], i), R[c[i]] = max(R[c[i]], i + 1),
    Mn[i][i + 1] = c[i];
  for (register int l = (1); l <= (m); ++l)
    for (register int r = (l + 2); r <= (m + 1); ++r)
      Mn[l][r] = min(Mn[l][r - 1], c[r - 1]);
  printf("%d\n", F(1, m + 1).res);
  return 0;
}