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;
template <class A, class B>
inline bool mina(A &first, B second) {
  return (first > second) ? (first = second, 1) : 0;
}
template <class A, class B>
inline bool maxa(A &first, B second) {
  return (first < second) ? (first = second, 1) : 0;
}
template <typename Int, Int mod>
class ModInt {
 public:
  static const Int Mod = mod;
  Int first;
  ModInt() { first = 0; }
  ModInt(int a) {
    Int t = a % mod;
    if (t < 0) t += mod;
    first = t;
  }
  ModInt(long long a) {
    Int t = a % mod;
    if (t < 0) t += mod;
    first = t;
  }
  Int get() const { return first; }
  ModInt &operator+=(ModInt that) {
    if ((first += that.first) >= mod) first -= mod;
    return *this;
  }
  ModInt &operator-=(ModInt that) {
    if ((first += mod - that.first) >= mod) first -= mod;
    return *this;
  }
  ModInt &operator*=(ModInt that) {
    first = (long long)(first)*that.first % mod;
    return *this;
  }
  bool operator==(ModInt that) const { return first == that.first; }
  ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
  ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
  ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
  ModInt operator-() const { return ModInt(-this->first); }
  inline friend ostream &operator<<(ostream &out, ModInt m) {
    return out << m.first;
  }
  ModInt power(long long k) const {
    ModInt r(1);
    ModInt b = *this;
    if (k <= 0) return r;
    while (k) {
      if (k & 1) r *= b;
      b *= b;
      k >>= 1;
    }
    return r;
  }
  ModInt inverse() const {
    long long a = first, b = mod, u = 1, v = 0;
    while (b) {
      long long t = a / b;
      a -= t * b;
      swap(a, b);
      u -= t * v;
      swap(u, v);
    }
    return ModInt(u);
  }
};
const int MAXN = 1e6 + 5;
const int MAXC = 505;
int C, N;
map<int, ModInt<int, (998244353)> > dp[MAXN];
int left_p[MAXC];
int right_p[MAXC];
vector<int> V;
vector<int> adj[MAXC];
ModInt<int, (998244353)> solve(int left, int right) {
  if (left >= right) return 1;
  if (dp[left].count(right)) return dp[left][right];
  dp[left][right] = 0;
  int min_c = V[left];
  for (int i = left; i < right; i++) {
    mina(min_c, V[i]);
    i = right_p[V[i]];
  }
  ModInt<int, (998244353)> from_right = 0;
  for (int r = right_p[min_c]; r < right;) {
    from_right += solve(r + 1, right) * solve(right_p[min_c] + 1, r + 1);
    if (r + 1 < right) {
      r = right_p[V[r + 1]];
    } else {
      break;
    }
  }
  ModInt<int, (998244353)> temp2 = 1;
  int pre = left_p[min_c];
  for (int i = left_p[min_c] + 1; i <= right_p[min_c]; i++) {
    if (V[i] == min_c) {
      temp2 *= solve(pre + 1, i);
      pre = i;
    }
  }
  for (int l = left_p[min_c]; l >= left;) {
    ModInt<int, (998244353)> temp =
        from_right * solve(left, l) * solve(l, left_p[min_c]);
    dp[left][right] += temp;
    if (l - 1 >= left) {
      l = left_p[V[l - 1]];
    } else {
      break;
    }
  }
  dp[left][right] *= temp2;
  return dp[left][right];
}
int main() {
  cin >> C >> N;
  for (int i = 0; (i) < (N); ++(i)) {
    int c;
    cin >> c;
    c--;
    if (!V.empty() && V.back() == c) continue;
    V.push_back(c);
  }
  N = ((int)(V).size());
  for (int i = 0; (i) < (C); ++(i)) {
    left_p[i] = MAXN;
    right_p[i] = -1;
  }
  for (int i = 0; (i) < (N); ++(i)) {
    int c = V[i];
    mina(left_p[c], i);
    maxa(right_p[c], i);
  }
  for (int i = 0; (i) < (C); ++(i)) {
    for (int j = 0; (j) < (i); ++(j)) {
      if (left_p[i] <= left_p[j] && left_p[j] <= right_p[i] &&
          right_p[j] >= right_p[i]) {
        cout << 0 << endl;
        return 0;
      }
      if (left_p[j] <= left_p[i] && left_p[i] <= right_p[j] &&
          right_p[i] >= right_p[j]) {
        cout << 0 << endl;
        return 0;
      }
      if (left_p[i] <= left_p[j] && left_p[j] <= right_p[i] &&
          right_p[j] <= right_p[i]) {
        if (j < i) {
          cout << 0 << endl;
          return 0;
        } else {
          adj[i].push_back(j);
        }
      }
      if (left_p[j] <= left_p[i] && left_p[i] <= right_p[j] &&
          right_p[i] <= right_p[j]) {
        if (i < j) {
          cout << 0 << endl;
          return 0;
        } else {
          adj[j].push_back(i);
        }
      }
    }
  }
  cout << solve(0, N) << endl;
  return 0;
}