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;
int dp[505][505], N, M, arr[1000005], values[1000005];
int f(int x, int y) {
  if (x == y)
    return 1;
  else if (x > y)
    return 1;
  else if (dp[x][y] != -1)
    return dp[x][y];
  int mini = 1000000000, loc = 0;
  for (int i = x; i <= y; i++) {
    if (arr[i] < mini) {
      mini = arr[i];
      loc = i;
    }
  }
  int sum_le = 0, sum_re = 0;
  vector<int> le, ri;
  for (int l = x; l <= loc; l++) {
    long long val = f(x, l - 1);
    val *= f(l, loc - 1);
    val %= 998244353;
    le.push_back(val);
    sum_le += val;
    sum_le %= 998244353;
  }
  for (int r = loc; r <= y; r++) {
    long long val = f(loc + 1, r);
    val *= f(r + 1, y);
    val %= 998244353;
    ri.push_back(val);
    sum_re += val;
    sum_re %= 998244353;
  }
  long long ans = (long long)sum_le * sum_re;
  ans %= 998244353;
  return dp[x][y] = ans;
}
int compute(vector<int> arr1) {
  for (int i = 0; i < arr1.size(); i++) arr[i] = arr1[i];
  for (int i = 0; i < arr1.size(); i++)
    for (int j = 0; j < arr1.size(); j++) dp[i][j] = -1;
  return f(0, arr1.size() - 1);
}
vector<int> locations[505];
struct node {
  int s, e, m, mx;
  node *l, *r;
  node(int _s, int _e) {
    s = _s;
    e = _e;
    m = (s + e) / 2;
    if (s == e)
      mx = values[s];
    else {
      l = new node(s, m);
      r = new node(m + 1, e);
      mx = min(l->mx, r->mx);
    }
  }
  int qry(int qs, int qe) {
    if (qs > qe)
      return 1000000000;
    else if (qs <= s && e <= qe)
      return mx;
    else if (qs > e || s > qe)
      return 1000000000;
    else
      return min(l->qry(qs, qe), r->qry(qs, qe));
  }
} * root;
int fir[505], las[505];
long long ans = 1;
void decomp(int s, int e) {
  if (s > e) return;
  vector<int> res;
  for (int i = s; i <= e;) {
    res.push_back(values[i]);
    for (int j = 0; j < locations[values[i]].size() - 1; j++)
      decomp(locations[values[i]][j] + 1, locations[values[i]][j + 1] - 1);
    i = las[values[i]] + 1;
  }
  ans *= compute(res);
  ans %= 998244353;
}
int main() {
  scanf("%d%d", &M, &N);
  for (int i = 0; i < N; i++) scanf("%d", &values[i]);
  for (int i = 0; i < N; i++) locations[values[i]].push_back(i);
  root = new node(0, N - 1);
  memset(fir, 63, sizeof(fir));
  memset(las, -63, sizeof(las));
  for (int i = 0; i < N; i++) fir[values[i]] = min(fir[values[i]], i);
  for (int i = 0; i < N; i++) las[values[i]] = max(las[values[i]], i);
  bool flag = 0;
  for (int i = 1; i <= M; i++) {
    if (root->qry(fir[i], las[i]) < i) flag = 1;
  }
  if (flag) {
    printf("0\n");
    return 0;
  }
  decomp(0, N - 1);
  printf("%lld\n", ans);
}