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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
long long MOD = 998244353;
long double EPS = 1e-9;
long long binpow(long long b, long long p, long long mod) {
  long long ans = 1;
  b %= mod;
  for (; p; p >>= 1) {
    if (p & 1) ans = ans * b % mod;
    b = b * b % mod;
  }
  return ans;
}
void pre() {}
vector<long long> arr;
long long minpresent[501][501];
long long memo1[501][501];
long long memo2[501][501];
long long pos[501];
long long dp1(int i, int j);
long long dp2(int i, int j) {
  if (i > j) return 1;
  if (memo2[i][j] == -1) {
    long long ans = (2 * dp1(i, j)) % MOD;
    for (long long k = i; k <= j - 1; ++k) {
      ans = (ans + dp1(i, k) * dp1(k + 1, j)) % MOD;
    }
    memo2[i][j] = ans;
  }
  return memo2[i][j];
}
long long dp1(int i, int j) {
  if (i > j) return 0;
  if (memo1[i][j] == -1) {
    long long mini = pos[minpresent[i][j]];
    memo1[i][j] = dp2(i, mini - 1) * dp2(mini + 1, j) % MOD;
  }
  return memo1[i][j];
}
void solve() {
  long long n, m;
  cin >> n >> m;
  arr.resize(n);
  for (long long i = 0; i < (n); ++i) {
    cin >> arr[i];
    pos[arr[i]] = i;
  }
  for (long long lef = 0; lef <= n - 1; ++lef) {
    long long mini = MOD;
    for (long long rig = lef; rig <= n - 1; ++rig) {
      (mini = min((mini), (arr[rig])));
      minpresent[lef][rig] = mini;
    }
  }
  memset(memo1, -1, sizeof(memo1));
  memset(memo2, -1, sizeof(memo2));
  cout << dp1(0, n - 1) << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  pre();
  long long _t = 1;
  for (long long i = 1; i <= _t; i++) {
    solve();
  }
}