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;
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
int n, m;
int C[555];
int inv[555];
int L[555], R[555];
const int MOD = 998244353;
map<pair<int, int>, long long> dp[555];
long long p1sum[555][555];
long long p2sum[555][555];
long long count(int num, int l, int r);
long long cp1(int num, int l) {
  long long& ans = p1sum[num][l];
  int contains = inv[num];
  if (ans == -1) {
    ans = 0;
    int minivalLEFT = n + 10;
    set<int> L;
    for (int a = l; a < contains; a++) {
      L.insert(C[a]);
    }
    for (int a = l; a <= contains; a++) {
      int minival = n + 10;
      long long part1 = 1;
      if (L.size()) {
        part1 = count(*L.begin(), a, contains - 1);
      }
      if (a < contains) L.erase(C[a]);
      long long part2 = 1;
      if (minivalLEFT != n + 10) {
        part2 = count(minivalLEFT, l, a - 1);
      }
      if (C[a] < minivalLEFT && C[a] > num) minivalLEFT = C[a];
      ans = (ans + (part1 * part2 % MOD)) % MOD;
    }
  }
  return ans;
}
long long cp2(int num, int r) {
  long long& ans = p2sum[num][r];
  int contains = inv[num];
  if (ans == -1) {
    ans = 0;
    int minivalRIGHT = n + 10;
    set<int> R;
    for (int a = contains + 1; a <= r; a++) {
      R.insert(C[a]);
    }
    for (int b = r; b >= contains; b--) {
      long long part3 = 1;
      if (minivalRIGHT != n + 10) {
        part3 = count(minivalRIGHT, b + 1, r);
      }
      if (C[b] < minivalRIGHT && C[b] > num) minivalRIGHT = C[b];
      long long part4 = 1;
      if (R.size()) {
        part4 = count(*R.begin(), contains + 1, b);
      }
      ans = (ans + (part3 * part4 % MOD)) % MOD;
      if (b > contains) R.erase(C[b]);
    }
  }
  return ans;
}
long long count(int num, int l, int r) {
  if (dp[num].end() == dp[num].find(make_pair(l, r))) {
    dp[num][make_pair(l, r)] = 0;
    long long& ans = dp[num][make_pair(l, r)];
    int contains = inv[num];
    ans = 0;
    long long p1sumS = cp1(num, l);
    long long p2sumS = cp2(num, r);
    ans = p1sumS * p2sumS % MOD;
  }
  return dp[num][make_pair(l, r)];
}
int main() {
  memset(p1sum, -1, sizeof(p1sum));
  memset(p2sum, -1, sizeof(p2sum));
  scanf("%d%d", &n, &m);
  for (int(i) = 0; (i) < (m); (i)++) {
    scanf("%d", C + i);
    C[i]--;
    inv[C[i]] = i;
  }
  printf("%lld\n", count(0, 0, n - 1));
  return 0;
}