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 MAXN = (int)1e6 + 3;
const int MAXK = (int)1e3 + 3;
const int infint = (int)1e9 + 3;
const int MOD = (int)998244353;
const int LOG = (int)20;
const long long inf = (long long)1e17;
vector<int> pos[MAXK];
int add(int a, int b) {
  int c = (a + b);
  if (c >= MOD) c -= MOD;
  if (c < 0) c += MOD;
  return c;
}
int mul(int a, int b) {
  int c = 1LL * a * b % MOD;
  return c;
}
int n, m, a[MAXN], mn[LOG][MAXN], lg[MAXN];
int rmq(int l, int r) {
  int len = lg[r - l + 1];
  int ans = min(mn[len][l], mn[len][r - (1 << len) + 1]);
  return ans;
}
int l[MAXK], r[MAXK], tmp[MAXN], dp[MAXK][MAXK], pre[MAXK][MAXK];
bool valid() {
  for (int i = 0; i < m; i++) r[a[i]] = i;
  for (int i = m - 1; i >= 0; i--) l[a[i]] = i;
  for (int i = 1; i <= n; i++) {
    for (int j = l[i]; j <= r[i]; j++) tmp[j] = i;
  }
  for (int i = 0; i < m; i++)
    if (tmp[i] != a[i]) return 0;
  return 1;
}
int hsh[MAXK], rev[MAXN];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; i++) cin >> a[i];
  for (int i = 0; i < m; i++) pos[a[i]].push_back(i);
  lg[1] = 0;
  for (int i = 2; i < MAXN; i++) lg[i] = lg[i >> 1] + 1;
  memset(mn, -1, sizeof mn);
  for (int i = 0; i < m; i++) mn[0][i] = a[i];
  for (int i = 1; i < LOG; i++)
    for (int j = 0; j < m; j++)
      if (mn[i - 1][j] != -1)
        mn[i][j] = min(mn[i - 1][j], mn[i - 1][j + (1 << (i - 1))]);
  if (!valid()) return cout << 0, 0;
  vector<int> all;
  for (int i = 1; i <= n; i++) all.push_back(l[i]), all.push_back(r[i]);
  sort(all.begin(), all.end());
  all.resize(unique(all.begin(), all.end()) - all.begin());
  for (int i = 0; i < all.size(); i++) hsh[i] = all[i], rev[all[i]] = i;
  int sze = all.size();
  for (int i = 0; i < sze; i++) {
    int mka = a[hsh[i]];
    dp[i][i] = (l[mka] == hsh[i]) & (r[mka] == hsh[i]);
    pre[i][i] = 2 * dp[i][i];
  }
  for (int sz = 1; sz < sze; sz++)
    for (int i = 0; i + sz < sze; i++) {
      int j = i + sz;
      int idx = rmq(hsh[i], hsh[j]), L = rev[l[idx]], R = rev[r[idx]];
      if (L < i || R > j)
        dp[i][j] = 0;
      else {
        int ans = 1;
        if (L != i) ans = mul(ans, pre[i][L - 1]);
        if (R != j) ans = mul(ans, pre[R + 1][j]);
        for (int k = 1; k < pos[idx].size(); k++) {
          int xl = pos[idx][k - 1], xr = pos[idx][k];
          xl++, xr--;
          if (xl <= xr) ans = mul(ans, dp[rev[xl]][rev[xr]]);
        }
        dp[i][j] = ans;
      }
      pre[i][j] = dp[i][j];
      for (int x = i; x <= j; x++) {
        if (x == j)
          pre[i][j] = add(pre[i][j], dp[i][j]);
        else
          pre[i][j] = add(pre[i][j], mul(dp[i][x], dp[x + 1][j]));
      }
    }
  cout << dp[0][sze - 1];
}