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;
const int INF = 2147483647;
const long long INF2 = 9223372036854775807LL;
const double PI = acos(-1);
const double ERR = 1e-10;
const int N = 1000005;
long long n, m;
long long a[505];
long long ra[505];
long long dp[505][505];
namespace MIN_ver {
int INIT = 1000000000;
int node_calc(int n1, int n2) { return min(n1, n2); }
}  // namespace MIN_ver
namespace MAX_ver {
int INIT = 0;
int node_calc(int n1, int n2) { return max(n1, n2); }
}  // namespace MAX_ver
namespace SUM_ver {
int INIT = 0;
int node_calc(int n1, int n2) { return n1 + n2; }
}  // namespace SUM_ver
using namespace MIN_ver;
struct segment_tree {
  int tree_n;
  vector<int> tree;
  void init(vector<int>& v) {
    tree.resize(4 * v.size() + 1);
    for (int i = 0; i < tree.size(); i++) tree[i] = INIT;
    tree_n = 1;
    while (tree_n < v.size()) tree_n <<= 1;
    for (int i = 0; i < v.size(); i++) tree[tree_n + i - 1] = v[i];
    for (int i = tree_n - 2; i >= 0; i--)
      tree[i] = node_calc(tree[i * 2 + 1], tree[i * 2 + 2]);
  }
  void init(int vn) {
    vector<int> v(vn + 1, INIT);
    init(v);
  }
  void recalc(int position) {
    while (position != 0) {
      position = (position - 1) >> 1;
      tree[position] =
          node_calc(tree[position * 2 + 1], tree[position * 2 + 2]);
    }
  }
  void update_push(int position, int v) {
    position += tree_n - 1;
    tree[position] = v;
    recalc(position);
  }
  void update_calc(int position, int v) {
    position += tree_n - 1;
    tree[position] = node_calc(tree[position], v);
    recalc(position);
  }
  int query(int a, int b, int idx, int left, int right) {
    if (right <= a || left >= b) return INIT;
    if (left >= a && right <= b) return tree[idx];
    int mid = (left + right) / 2;
    return node_calc(query(a, b, idx * 2 + 1, left, mid),
                     query(a, b, idx * 2 + 2, mid, right));
  }
  int query(int a, int b) { return query(a, b, 0, 0, tree_n); }
} smtree;
int main() {
  scanf("%lld", &n);
  scanf("%lld", &m);
  smtree.init(n + 5);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    ra[a[i]] = i;
    smtree.update_push(i, a[i]);
  }
  for (int i = 1; i <= n; i++) {
    dp[i][i + 1] = 1;
    dp[i][i] = 1;
  }
  dp[n + 1][n + 1] = 1;
  for (int o = 2; o <= n; o++) {
    for (int i = 1; i + o <= n + 1; i++) {
      int j = i + o;
      int idx = smtree.query(i, j);
      idx = ra[idx];
      long long s1 = 0;
      for (int k = idx; k >= i; k--) {
        s1 = (s1 + (dp[k][idx] * dp[i][k]) % MOD) % MOD;
      }
      long long s2 = 0;
      int id1 = idx + 1;
      for (int k = id1; k <= j; k++) {
        s2 = (s2 + (dp[id1][k] * dp[k][j]) % MOD) % MOD;
      }
      dp[i][j] = (s1 * s2) % MOD;
    }
  }
  printf("%lld", dp[1][n + 1] % MOD);
  return 0;
}