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("O3")
using namespace std;
const int N = 505;
long long dp[N][N], ans[N][N];
int a[N], loc[N];
inline long long mul(long long x, long long y) { return x * y % 998244353; }
inline long long add(long long x, long long y) {
  long long res = x + y;
  if (res >= 998244353) res -= 998244353;
  return res;
}
inline long long sub(long long x, long long y) {
  long long res = x + 998244353 - y;
  if (res >= 998244353) res -= 998244353;
  return res;
}
long long fac[N], inv[N];
long long mod_pow(long long x, long long p) {
  long long res = 1;
  while (p > 0) {
    if (p & 1) res = res * x % 998244353;
    p >>= 1;
    x = x * x % 998244353;
  }
  return res;
}
inline long long comb(long long n, long long k) {
  long long res = mul(fac[n], inv[k]);
  res = mul(res, inv[n - k]);
  return res;
}
struct solver {
  long long gans(int l, int r) {
    if (l >= r) return 1;
    if (!ans[l][r]) {
      int bestv = INT_MAX, bestind = -1;
      for (int i = l; i <= (int)r; ++i) {
        if (a[i] < bestv) {
          bestv = a[i];
          bestind = i;
        }
      }
      ans[l][r] = mul(gdp(l, bestind - 1), gdp(bestind + 1, r));
    }
    return ans[l][r];
  }
  long long gdp(int l, int r) {
    if (l > r) return 1;
    if (l == r) return 2;
    if (!dp[l][r]) {
      for (int i = l - 1; i <= (int)r; ++i) {
        dp[l][r] = add(dp[l][r], mul(gans(l, i), gans(i + 1, r)));
      }
    }
    return dp[l][r];
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= (int)n; ++i) {
    cin >> a[i];
    loc[a[i]] = i;
  }
  for (int i = 1; i <= (int)n; ++i) dp[i][i] = 1;
  for (int i = 1; i <= (int)n; ++i) ans[i][i] = 1;
  fac[0] = inv[0] = 1;
  for (int i = 1; i <= (int)n; ++i) {
    fac[i] = mul(fac[i - 1], i);
    inv[i] = mod_pow(fac[i], 998244353 - 2);
  }
  solver solve;
  long long res = solve.gans(1, n);
  cout << res << '\n';
}