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 long long MOD = 998244353;
struct intmod {
  long long val;
  intmod(long long _val = 1) {
    val = _val % MOD;
    if (val < 0) val += MOD;
  }
  bool operator!=(intmod rhs) const { return val != rhs.val; }
  bool operator==(intmod rhs) const { return val == rhs.val; }
  intmod operator+=(intmod rhs) {
    val += rhs.val;
    if (val >= MOD) val -= MOD;
    return *this;
  }
  intmod operator-=(intmod rhs) {
    val -= rhs.val;
    if (val < 0) val += MOD;
    return *this;
  }
  intmod operator*=(intmod rhs) {
    val = val * rhs.val % MOD;
    return *this;
  }
  friend intmod operator+(intmod lhs, intmod rhs) { return lhs += rhs; }
  friend intmod operator-(intmod lhs, intmod rhs) { return lhs -= rhs; }
  friend intmod operator*(intmod lhs, intmod rhs) { return lhs *= rhs; }
  friend intmod sqr(intmod a) { return a * a; }
  friend intmod bpow(intmod a, int p) {
    intmod ans(1);
    while (p) {
      if (p & 1) ans *= a;
      a = sqr(a);
      p >>= 1;
    }
    return ans;
  }
  friend intmod inv(intmod a) { return bpow(a, MOD - 2); }
  intmod operator/=(intmod rhs) { return (*this) *= inv(rhs); }
  friend intmod operator/(intmod lhs, intmod rhs) { return lhs /= rhs; }
  friend istream &operator>>(istream &is, intmod &im) { return is >> im.val; }
  friend ostream &operator<<(ostream &os, intmod im) { return os << im.val; }
};
const int MAXN = 517, MAXM = 1000030, LOG = 21;
intmod F0[MAXN][MAXN], F1[MAXN][MAXN];
intmod solve(vector<int> vec) {
  int N = ((int)vec.size());
  vec.insert(vec.begin(), MAXN);
  for (int i = 0; i <= N + 1; ++i)
    for (int j = 0; j <= N + 1; ++j) F0[i][j] = F1[i][j] = 1;
  for (int i = 1; i <= N; ++i) F0[i][i] = F1[i][i] = 1;
  for (int len = 2; len <= N; ++len)
    for (int l = 1; l <= N - len + 1; ++l) {
      int r = l + len - 1, m = l;
      for (int i = l + 1; i <= r; ++i)
        if (vec[i] < vec[m]) m = i;
      F0[l][r] = F1[l][m - 1] * F1[m + 1][r];
      F1[l][r] = 0;
      for (int i = l; i <= r; ++i) F1[l][r] += F0[l][i] * F1[i + 1][r];
    }
  return F1[1][N];
}
int N, M;
int A[MAXM], L[MAXN], R[MAXN];
int st[MAXN], rr = 0;
vector<int> vec[MAXN * 2];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  scanf("%d %d", &N, &M);
  for (int i = 1; i <= M; ++i) scanf("%d", &A[i]);
  for (int i = 1; i <= M; ++i) {
    R[A[i]] = i;
    if (!L[A[i]]) L[A[i]] = i;
  }
  int ptr = N + 1;
  for (int i = 1; i <= M; ++i) {
    int last = (rr > 0 ? st[rr - 1] : 0);
    if (last > A[i]) {
      puts("0");
      return 0;
    }
    if (L[A[i]] == i) {
      vec[last].push_back(A[i]);
      st[rr++] = A[i];
    }
    if (A[i] == last && !vec[last].empty()) vec[ptr++].swap(vec[last]);
    if (R[A[i]] == i) --rr;
  }
  assert(ptr < MAXN * 2);
  intmod ans = 1;
  for (int i = 0; i <= ptr; ++i)
    if (!vec[i].empty()) ans *= solve(vec[i]);
  cout << ans << endl;
  return 0;
}