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;
const int P = 998244353;
const int N = 1000010;
int n, m;
int c[N];
vector<int> colors[510];
int ans = 1;
int dp[510][510];
int l[510], r[510];
int order_l[510], order_r[510];
int tmpl[510], tmpr[510];
int f[N];
vector<pair<int, int>> v[N];
bool cmpl(int x, int y) { return l[x] < l[y]; }
bool cmpr(int x, int y) { return r[x] < r[y]; }
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d", &c[i]);
    colors[c[i]].push_back(i);
  }
  set<int> id;
  id.insert(0);
  id.insert(m + 1);
  for (int i = 1; i <= n; i++) {
    assert(!colors[i].empty());
    int l = colors[i].front(), r = colors[i].back();
    auto it = id.lower_bound(l);
    if (*it <= r) {
      printf("0\n");
      return 0;
    }
    it--;
  }
  for (int i = 1; i <= n; i++)
    l[i] = colors[i].front(), r[i] = colors[i].back();
  for (int i = 1; i <= n; i++) order_l[i] = i;
  sort(order_l + 1, order_l + n + 1, cmpl);
  for (int i = 1; i <= n; i++) order_r[i] = i;
  sort(order_r + 1, order_r + n + 1, cmpr);
  l[0] = 0;
  r[0] = m + 1;
  for (int i = n; i; i--) {
    memset(tmpl, 0, sizeof tmpl);
    tmpl[i] = 1;
    for (int j = n; j; j--) {
      for (int k = i; k <= n; k++)
        if (l[c[r[k] + 1]] == r[k] + 1)
          tmpl[order_l[j]] =
              (tmpl[order_l[j]] + 1LL * tmpl[c[r[k] + 1]] * dp[order_l[j]][k]) %
              P;
    }
    memset(tmpr, 0, sizeof tmpr);
    tmpr[i] = 1;
    for (int j = 1; j <= n; j++) {
      for (int k = i; k <= n; k++)
        if (r[c[l[k] - 1]] == l[k] - 1)
          tmpr[order_r[j]] =
              (tmpr[order_r[j]] + 1LL * tmpr[c[l[k] - 1]] * dp[k][order_r[j]]) %
              P;
    }
    for (int j = i; j <= n; j++)
      for (int k = i; k <= n; k++)
        dp[j][k] = (dp[j][k] + 1LL * tmpl[j] * tmpr[k]) % P;
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      v[l[i] - 1].push_back(make_pair(r[j], dp[i][j]));
    }
  f[0] = 1;
  for (int i = 0; i <= m; i++) {
    for (auto p : v[i]) f[p.first] = (f[p.first] + 1LL * p.second * f[i]) % P;
  }
  printf("%d\n", f[m]);
}