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;
struct vec {
  long double x, y;
  bool operator==(const vec &a) const { return x == a.x && y == a.y; }
  bool operator<(const vec &a) const {
    if (a.x != x) return x < a.x;
    return y < a.y;
  }
  vec operator+(const vec &a) const { return {x + a.x, y + a.y}; }
  vec operator*(const long double &a) const { return {x * a, y * a}; }
  vec operator-(const vec &a) const { return {x - a.x, y - a.y}; }
  vec operator/(const long double &a) const { return {x / a, y / a}; }
  long double len() const { return sqrt(x * x + y * y); }
  long double dot(const vec &a) const { return x * a.x + y * a.y; }
  long double crs(const vec &a) const { return x * a.y - y * a.x; }
  vec proj(const vec &a) const {
    return (a / a.len()) * (*this).dot(a) / a.len();
  }
};
struct P {
  int x, y;
};
bool as(P a, P b) { return a.x < b.x; }
long long x, y, z, mod = 998244353;
vector<int> v;
int i, n, m, k, a, d, b, c;
int e;
int o[511];
long long l[511][511][4];
int j[33333];
stack<int> s;
int u[551][555];
string r;
queue<P> q;
long long g(int n, int m, int c) {
  if (n >= m) return 1;
  if (l[n][m][c]) return l[n][m][c];
  int d = u[n][m];
  if (c == 0) {
    long long x = g(n, m, 1);
    for (int t = n; t < d; t++)
      x += g(n, t, 0) * g(t + 1, m, 1) % mod, x %= mod;
    return l[n][m][c] = x;
  }
  if (c == 1) {
    long long x = g(n, m, 2);
    for (int t = m; t > d; t--)
      x += g(n, t - 1, 2) * g(t, m, 0) % mod, x %= mod;
    return l[n][m][c] = x;
  }
  return l[n][m][c] = g(n, d - 1, 0) * g(d + 1, m, 0) % mod;
}
int main() {
  scanf("%d %d", &a, &e);
  o[0] = 987654321;
  for (int t = 1; t <= a; t++) scanf("%d", &o[t]);
  for (int t = 1; t <= a; t++)
    for (int w = t; w <= a; w++) {
      int k = 987654321, n = 0;
      for (int h = t; h <= w; h++)
        if (o[h] < k) n = h, k = o[h];
      u[t][w] = n;
    }
  printf("%d", g(1, a, 0));
}