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 n, m, x[1000005], par[1000005];
long long ans;
vector<int> v[505];
vector<long long> dp[505];
void add(long long& a, long long b) {
  a += b;
  if (a >= 998244353) a -= 998244353;
}
int find(int a) { return par[a] == a ? a : par[a] = find(par[a]); }
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d", x + i);
    v[x[i]].push_back(i);
  }
  for (int i = n; i; i--) {
    auto& w = v[i];
    par[w[0]] = w[0];
    long long mul = 1;
    for (int j = 1; j < w.size(); j++) {
      int at = w[j];
      if (par[at - 1] == 0) {
        puts("0");
        return 0;
      }
      int tmp = find(at - 1);
      if (tmp != w[0]) {
        if (par[tmp - 1] == 0 || find(tmp - 1) != w[0]) {
          puts("0");
          return 0;
        }
        par[tmp] = w[0];
        long long sum = 0;
        for (auto k : dp[x[tmp]]) sum += k;
        mul = sum % 998244353 * mul % 998244353;
      }
      par[at] = w[0];
    }
    int cur = w[0];
    if (par[w[0] - 1]) {
      cur = find(w[0] - 1);
      par[w[0]] = cur;
      auto& ww = dp[x[cur]];
      vector<long long> tdp(ww.size() + 1);
      for (int j = 0; j < ww.size(); j++) {
        tdp[j + 1] = tdp[j];
        add(tdp[j + 1], ww[j]);
      }
      ww.push_back(0);
      for (int j = 0; j < ww.size(); j++) {
        ww[j] = tdp.back();
        add(ww[j], 998244353 - tdp[max(0, j - 1)]);
      }
    } else
      dp[x[cur]] = {1};
    if (par[w.back() + 1]) {
      int tmp = w.back() + 1;
      par[tmp] = cur;
      auto& ww = dp[x[tmp]];
      vector<long long> tdp(ww.size() + 1);
      for (int j = 0; j < ww.size(); j++) {
        tdp[j + 1] = tdp[j];
        add(tdp[j + 1], ww[j]);
      }
      ww.push_back(0);
      for (int j = 0; j < ww.size(); j++) {
        ww[j] = tdp.back();
        add(ww[j], 998244353 - tdp[max(0, j - 1)]);
      }
      auto& www = dp[x[cur]];
      auto ndp = vector<long long>(ww.size() - 1 + www.size());
      for (int k = 0; k < ww.size(); k++)
        for (int j = 0; j < www.size(); j++)
          ndp[k + j] = (ndp[k + j] + ww[k] * www[j]) % 998244353;
      www.swap(ndp);
    }
    for (auto& j : dp[x[cur]]) j = j * mul % 998244353;
  }
  long long sum = 0;
  for (auto i : dp[x[1]]) sum += i;
  printf("%lld\n", sum % 998244353);
}