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, k;
int init[1000007];
struct tuhla {
  int col, len;
};
tuhla a[1007];
int dp[1007][1007];
vector<int> prec[1007];
int mnval[1007][1007];
int st[1007];
int en[1007];
bool check() {
  set<int> s;
  for (int i = 1; i <= n; ++i) {
    prec[init[i]].push_back(i);
  }
  for (int i = 1; i <= k; ++i) {
    auto it = s.upper_bound(prec[i].front());
    if (it != s.end() && (*it) < prec[i].back()) {
      return false;
    }
    for (auto x : prec[i]) {
      s.insert(x);
    }
  }
  return true;
}
void input() {
  scanf("%d%d", &k, &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &init[i]);
  }
}
void solve() {
  if (check() == false) {
    printf("0\n");
    return;
  }
  int cnt = 1;
  int dummy = n;
  n = 0;
  for (int i = 2; i <= dummy; ++i) {
    if (init[i] == init[i - 1]) {
      ++cnt;
    } else {
      a[++n] = (tuhla){init[i - 1], cnt};
      cnt = 1;
    }
  }
  a[++n] = (tuhla){init[dummy], cnt};
  for (int i = 1; i <= n; ++i) {
    mnval[i][i] = a[i].col;
    for (int j = i + 1; j <= n; ++j) {
      mnval[i][j] = min(mnval[i][j - 1], a[j].col);
    }
  }
  for (int i = 1; i <= n; ++i) {
    en[a[i].col] = i;
  }
  for (int i = n; i >= 1; --i) {
    st[a[i].col] = i;
  }
  for (int i = 1; i <= k; ++i) {
    prec[i].clear();
  }
  for (int i = 1; i <= n; ++i) {
    prec[a[i].col].push_back(i);
  }
  for (int i = 0; i <= n; ++i) {
    for (int j = 0; j < i; ++j) {
      dp[i][j] = 1;
    }
  }
  for (int diff = 0; diff < n; ++diff) {
    for (int l = 1; l + diff <= n; ++l) {
      int r = l + diff;
      int sr = mnval[l][r];
      if (st[sr] < l || r < en[sr]) {
        continue;
      }
      int coef = 1;
      int sz = prec[sr].size();
      for (int i = 0; i < sz - 1; ++i) {
        coef =
            (1LL * coef * dp[prec[sr][i] + 1][prec[sr][i + 1] - 1]) % 998244353;
      }
      int aux1, aux2;
      aux1 = aux2 = 0;
      for (int t = l; t <= st[sr]; ++t) {
        int add = 1;
        if (l < t) {
          add = dp[l][t - 1];
        }
        if (t < st[sr]) {
          add = (1LL * add * dp[t][st[sr] - 1]) % 998244353;
        }
        aux1 += add;
        if (aux1 >= 998244353) {
          aux1 -= 998244353;
        }
      }
      for (int t = en[sr]; t <= r; ++t) {
        int add = 1;
        if (t < r) {
          add = dp[t + 1][r];
        }
        if (en[sr] < t) {
          add = (1LL * add * dp[en[sr] + 1][t]) % 998244353;
        }
        aux2 += add;
        if (aux2 >= 998244353) {
          aux2 -= 998244353;
        }
      }
      dp[l][r] = coef;
      dp[l][r] =
          (1LL * dp[l][r] * ((1LL * aux1 * aux2) % 998244353)) % 998244353;
    }
  }
  printf("%d\n", dp[1][n]);
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    input();
    solve();
  }
  return 0;
}