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;
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  bool first = true;
  for (const auto& j : v) {
    if (!first) {
      os << " ";
    }
    first = false;
    os << j;
  }
  os << "\n";
  return os;
}
template <typename T>
istream& operator>>(istream& is, vector<T>& v) {
  for (auto& elem : v) {
    is >> elem;
  }
  return is;
}
class segment_tree_max {
 private:
  vector<int> m;
  vector<int> delta;
  int n;
  bool is_bulk;
  int get_val(int i) {
    if (!is_bulk) {
      return m[i];
    }
    int ans = m[i] + delta[i];
    i /= 2;
    while (i >= 1) {
      ans += delta[i];
      i /= 2;
    }
    return ans;
  }
  void inc(int i, int val) {
    delta[i] += val;
    i /= 2;
    while (i >= 1) {
      m[i] = max(m[2 * i] + delta[2 * i], m[2 * i + 1] + delta[2 * i + 1]);
      i /= 2;
    }
  }

 public:
  segment_tree_max() {}
  segment_tree_max(int nn, int def, bool bulk = false) {
    is_bulk = bulk;
    n = 1;
    while (n < nn) {
      n *= 2;
    }
    m.assign(2 * n, def);
    if (is_bulk) {
      delta.assign(2 * n, 0);
    }
  }
  segment_tree_max(const vector<int>& d, int def = 0, bool bulk = false,
                   int mult = 1) {
    is_bulk = bulk;
    n = 1;
    while (n < d.size()) {
      n *= 2;
    }
    m.assign(2 * n, def);
    if (is_bulk) {
      delta.assign(2 * n, 0);
    }
    for (int i = 0; i < d.size(); i++) {
      m[i + n] = d[i] * mult;
    }
    for (int i = n - 1; i >= 1; i--) {
      m[i] = max(m[2 * i], m[2 * i + 1]);
    }
  }
  int get_max(int l, int r) {
    if (l > r) {
      swap(l, r);
    }
    l += n;
    r += n;
    int ans = -2e9;
    while (l <= r) {
      if (l % 2 == 1) {
        ans = max(ans, get_val(l));
        l++;
      }
      if (r % 2 == 0) {
        ans = max(ans, get_val(r));
        r--;
      }
      l /= 2;
      r /= 2;
    }
    return ans;
  }
  void inc_bulk(int l, int r, int val) {
    if (!is_bulk) {
      cout << "cant inc_bulk in not bulk mode" << endl;
      exit(1);
    }
    if (l > r) {
      swap(l, r);
    }
    l += n;
    r += n;
    while (l <= r) {
      if (l % 2 == 1) {
        inc(l, val);
        l++;
      }
      if (r % 2 == 0) {
        inc(r, val);
        r--;
      }
      l /= 2;
      r /= 2;
    }
  }
  void update(int i, int val) {
    i += n;
    m[i] = val;
    i /= 2;
    while (i >= 1) {
      if (is_bulk) {
        m[i] = max(m[2 * i] + delta[2 * i], m[2 * i + 1] + delta[2 * i + 1]);
      } else {
        m[i] = max(m[2 * i], m[2 * i + 1]);
      }
      i /= 2;
    }
  }
};
class segment_tree_min {
 private:
  segment_tree_max max_tree;

 public:
  segment_tree_min() {}
  segment_tree_min(int nn, int def = 2e9, bool bulk = false) {
    max_tree = segment_tree_max(nn, -def, bulk);
  }
  segment_tree_min(const vector<int>& d, int def = 2e9, bool bulk = false) {
    max_tree = segment_tree_max(d, -def, bulk, -1);
  }
  int get_min(int l, int r) { return -max_tree.get_max(l, r); }
  void inc_bulk(int l, int r, int val) { max_tree.inc_bulk(l, r, -val); }
  void update(int i, int val) { max_tree.update(i, -val); }
};
void solve() {
  int n, m;
  cin >> n >> m;
  vector<int> a(m);
  cin >> a;
  segment_tree_min segmentTreeMin(a);
  vector<int> mi2(n + 1, m + 1);
  vector<int> ma2(n + 1, 0);
  for (int i = 0; i < m; i++) {
    mi2[a[i]] = min(mi2[a[i]], i);
    ma2[a[i]] = max(ma2[a[i]], i);
  }
  for (int j = 1; j <= n; j++) {
    if (segmentTreeMin.get_min(mi2[j], ma2[j]) < j) {
      cout << 0;
      return;
    }
  }
  int j = 1;
  for (int i = 1; i < m; i++) {
    if (a[i] != a[i - 1]) {
      a[j++] = a[i];
    }
  }
  m = j;
  a.resize(m);
  mi2.assign(n + 1, m + 1);
  ma2.assign(n + 1, 0);
  for (int i = 0; i < m; i++) {
    mi2[a[i]] = min(mi2[a[i]], i);
    ma2[a[i]] = max(ma2[a[i]], i);
  }
  vector<vector<long long> > dp(m + 1, vector<long long>(m + 1, 1));
  for (int l = 2; l <= m; l++) {
    for (int j = 0; j <= m - l; j++) {
      for (int i = j; i < j + l; i++) {
        if (mi2[a[i]] < j || ma2[a[i]] >= j + l) {
          dp[j][l] = 0;
          break;
        }
      }
      if (dp[j][l] == 1) {
        int mi = j;
        int ma = j;
        for (int i = j; i < j + l; i++) {
          if (a[i] < a[mi]) {
            mi = i;
          }
          if (a[i] == a[mi]) {
            ma = i;
          }
        }
        long long x = 0;
        for (int i = j; i <= mi; i++) {
          x = (x + (dp[j][i - j] * dp[i][mi - i]) % 998244353) % 998244353;
        }
        long long y = 0;
        for (int i = ma; i < j + l; i++) {
          y = (y +
               (dp[ma + 1][i - ma] * dp[i + 1][j + l - 1 - i]) % 998244353) %
              998244353;
        }
        long long z = (x * y) % 998244353;
        for (int i = mi + 1; i <= ma; i++) {
          if (a[i] == a[mi]) {
            z = (z * dp[mi + 1][i - mi - 1]) % 998244353;
            mi = i;
          }
        }
        dp[j][l] = z;
      }
    }
  }
  cout << dp[0][m];
}
int main() {
  ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  if (0) {
    int tests;
    cin >> tests;
    while (tests--) {
      solve();
      cout << "\n";
    }
  } else {
    solve();
    cout << endl;
  }
  return 0;
}