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>
const int MOD = 998244353;
int cs[1000005];
int ds[1000005];
int ds_len = 0;
int first[1000][1000];
int dp[1000][1000];
std::vector<int> where[501];
int left[501];
int right[501];
int core[501];
int get_dp(int l, int r);
int get_core(int c) {
  if (core[c] != -1) return core[c];
  int res = 1;
  for (int i = 0; i + 1 < where[c].size(); i++) {
    int a = where[c][i], b = where[c][i + 1];
    res = 1LL * res * get_dp(a + 1, b - 1) % MOD;
  }
  core[c] = res;
  return core[c];
}
int get_dp(int l, int r) {
  if (l > r) return 1;
  if (dp[l][r] != -1) return dp[l][r];
  int c = first[l][r];
  int lsum = 0, rsum = 0;
  for (int a = l; a <= left[c]; a++) {
    lsum = (lsum + 1LL * get_dp(l, a - 1) * get_dp(a, left[c] - 1)) % MOD;
  }
  for (int b = right[c]; b <= r; b++) {
    rsum = (rsum + 1LL * get_dp(right[c] + 1, b) * get_dp(b + 1, r)) % MOD;
  }
  dp[l][r] = 1LL * lsum * get_core(c) % MOD * rsum % MOD;
  return dp[l][r];
}
int main() {
  int N, M;
  scanf("%d %d", &N, &M);
  for (int i = 0; i < M; i++) {
    scanf("%d", &cs[i]);
  }
  for (int i = 0; i < M; i++) {
    if (i == 0 || cs[i] != cs[i - 1]) {
      ds[ds_len++] = cs[i];
    }
  }
  if (ds_len > N * 2) {
    printf("0\n");
    return 0;
  }
  for (int i = 0; i < ds_len; i++) {
    first[i][i] = ds[i];
  }
  for (int l = 0; l < ds_len; l++) {
    for (int r = l + 1; r < ds_len; r++) {
      first[l][r] = std::min(first[l][r - 1], ds[r]);
    }
  }
  for (int i = 0; i < ds_len; i++) {
    where[ds[i]].push_back(i);
  }
  for (int c = 1; c <= N; c++) {
    assert(where[c].size());
    left[c] = where[c].front();
    right[c] = where[c].back();
  }
  for (int i = ds_len - 1; i >= 0; i--) {
    left[ds[i]] = i;
  }
  for (int i = 1; i <= N; i++) {
    if (first[left[i]][right[i]] < i) {
      printf("0\n");
      return 0;
    }
  }
  for (int i = 0; i < ds_len; i++) {
    for (int j = 0; j < ds_len; j++) {
      dp[i][j] = -1;
    }
  }
  for (int i = 1; i <= N; i++) {
    core[i] = -1;
  }
  printf("%d\n", get_dp(0, ds_len - 1));
}