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;
using ll = long long;
using ld = long double;
mt19937_64 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
const ld pi = acos(0) * 2;
template <typename T>
inline void read(T &x) {
  char c;
  bool nega = 0;
  while ((!isdigit(c = getchar())) && (c != '-'))
    ;
  if (c == '-') {
    nega = 1;
    c = getchar();
  }
  x = c - 48;
  while (isdigit(c = getchar())) x = x * 10 + c - 48;
  if (nega) x = -x;
}
template <typename T>
inline void writep(T x) {
  if (x > 9) writep(x / 10);
  putchar(x % 10 + 48);
}
template <typename T>
inline void write(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  writep(x);
}
template <typename T>
inline void writeln(T x) {
  write(x);
  putchar('\n');
}
template <typename CT, typename T>
inline void reset_container(CT &c, int sz, T v) {
  c.resize(sz);
  for (auto &x : c) x = v;
}
int n, m;
int c[1000001];
vector<int> a;
int first[501];
int last[501];
int Kafuu_Chino[2001];
int mi[2001][2001];
bool done[2001][2001];
ll f[2001][2001];
const ll base = 998244353;
ll F(int u, int v) {
  if (u > v) return 1;
  if (done[u][v]) return f[u][v];
  done[u][v] = 1;
  if (u == v) return f[u][v] = 1;
  int val = mi[u][v];
  ll left = 0, right = 0, mid = 1;
  {
    int x = first[val];
    while (x != last[val]) {
      mid = (mid * F(x + 1, Kafuu_Chino[x] - 1)) % base;
      x = Kafuu_Chino[x];
    }
  }
  {
    int near = 0;
    for (int i = (last[val]); i <= (v); i++) {
      near = max(near, last[a[i]]);
      if (i == near)
        right = (right + (F(last[val] + 1, i) * F(i + 1, v)) % base) % base;
    }
  }
  {
    int near = m + 1;
    for (int i = (first[val]); i >= (u); i--) {
      near = min(near, first[a[i]]);
      if (i == near)
        left = (left + (F(i, first[val] - 1) * F(u, i - 1)) % base) % base;
    }
  }
  return f[u][v] = (mid * ((left * right) % base)) % base;
}
int main() {
  read(n);
  read(m);
  for (int i = (1); i <= (m); i++) read(c[i]);
  a.push_back(0);
  for (int i = (1); i <= (m); i++)
    if (c[i] != a.back()) a.push_back(c[i]);
  m = a.size() - 1;
  for (int i = (1); i <= (m); i++)
    if (first[a[i]] == 0) first[a[i]] = i;
  for (int i = (m); i >= (1); i--)
    if (last[a[i]] == 0) last[a[i]] = i;
  for (int i = (1); i <= (n); i++) {
    for (int j = (first[i]); j <= (last[i]); j++) {
      if (a[j] < i) {
        writeln(0);
        return 0;
      }
    }
  }
  for (int i = (1); i <= (m); i++) mi[i][i] = a[i];
  for (int len = (2); len <= (m); len++) {
    for (int i = (1); i <= (m); i++) {
      int j = i + len - 1;
      if (j > m) break;
      mi[i][j] = min(mi[i][j - 1], mi[i + 1][j]);
    }
  }
  for (int i = (1); i <= (m); i++)
    for (int j = (i + 1); j <= (m); j++)
      if (a[i] == a[j]) {
        Kafuu_Chino[i] = j;
        break;
      }
  writeln(F(1, m));
}