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;
inline long long int modadd(long long int n, long long int m,
                            long long int p = 998244353) {
  return ((n + m) % p + p) % p;
}
inline long long int modsub(long long int n, long long int m,
                            long long int p = 998244353) {
  return ((n - m + p) % p + p) % p;
}
inline long long int modpro(long long int n, long long int m,
                            long long int p = 998244353) {
  return (((n % p) * (m % p)) % p + p) % p;
}
unsigned long long int powe(long long int first, long long int second) {
  unsigned long long int res = 1;
  while (second > 0) {
    if (second & 1) res = res * first;
    second = second >> 1;
    first = first * first;
  }
  return res;
}
long long int modpow(long long int first, long long int second,
                     long long int p = 998244353) {
  long long int res = 1;
  while (second > 0) {
    if (second & 1) res = modpro(res, first, p);
    second = second >> 1;
    first = modpro(first, first, p);
  }
  return res;
}
inline long long int modInverse(long long int n, long long int p = 998244353) {
  if (n == 1) return 1;
  return modpow(n, p - 2, p);
}
inline long long int moddiv(long long int n, long long int m,
                            long long int p = 998244353) {
  return modpro(n, modInverse(m, p), p);
}
inline long long int modadd3(long long int first, long long int second,
                             long long int z, long long int p = 998244353) {
  return modadd(modadd(first, second, p), z, p);
}
inline long long int modadd4(long long int first, long long int second,
                             long long int z, long long int w,
                             long long int p = 998244353) {
  return modadd(modadd(first, second, p), modadd(z, w, p), p);
}
inline long long int modnCr(long long int fac[], int n, int r,
                            long long int p = 998244353) {
  if (r == 0) return 1;
  return modpro(fac[n], modInverse(modpro(fac[r], fac[n - r], p), p), p);
}
template <typename T>
inline T max3(T first, T second, T z) {
  return max(max(first, second), z);
}
template <typename T>
inline T max4(T first, T second, T z, T w) {
  return max(max3(first, second, w), z);
}
template <typename T>
inline T min3(T first, T second, T z) {
  return min(min(first, second), z);
}
template <typename T>
inline T min4(T first, T second, T z, T w) {
  return min(min3(first, second, w), z);
}
template <typename T>
void printArr(T *arr, int s, int n) {
  for (int i = s; i <= n; i++) {
    cout << arr[i] << " ";
  }
  cout << endl;
}
int c[1000005];
long long int dp[2005][2005];
int j = 1;
long long int f(int l, int r) {
  if (l > r) return 1;
  long long int &res = dp[l][r];
  if (res != -1) return res;
  int lc = INT_MAX, poslc;
  for (int i = (l); i < (r + 1); i++) lc = min(lc, c[i]);
  for (int i = (0); i < (j); i++)
    if ((i < l || i > r) && c[i] == lc) return res = 0;
  vector<int> v;
  for (int i = (l); i < (r + 1); i++) {
    if (c[i] == lc) v.push_back(i);
  }
  long long int temp = 0;
  for (int i = (l); i < (v[0] + 1); i++) {
    temp = modadd(temp, modpro(f(l, i - 1), f(i, v[0] - 1)));
  }
  res = temp;
  temp = 0;
  for (int i = (v.back()); i < (r + 1); i++) {
    temp = modadd(temp, modpro(f(i + 1, r), f(v.back() + 1, i)));
  }
  res = modpro(res, temp);
  for (int i = (0); i < (((long long int)(v).size()) - 1); i++) {
    res = modpro(res, f(v[i] + 1, v[i + 1] - 1));
  }
  return res;
}
int ct[1000005];
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  int erer = 1;
  for (int erer2 = (1); erer2 < (erer + 1); erer2++) {
    int n, m;
    cin >> n >> m;
    for (int i = (0); i < (m); i++) cin >> ct[i];
    c[0] = ct[0];
    for (int i = (1); i < (m); i++)
      if (ct[i - 1] != ct[i]) c[j++] = ct[i];
    if (j > 2 * n) {
      cout << 0;
      return 0;
    }
    memset(dp, -1, sizeof(dp));
    cout << f(0, j - 1);
  }
  return 0;
}