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;
bool isrange(int second, int first, int n, int m) {
  if (0 <= second && second < n && 0 <= first && first < m) return true;
  return false;
}
int dy[4] = {1, 0, -1, 0}, dx[4] = {0, 1, 0, -1},
    ddy[8] = {1, 0, -1, 0, 1, 1, -1, -1}, ddx[8] = {0, 1, 0, -1, 1, -1, 1, -1};
const long long int mod = 998244353;
pair<int, int> len[555];
set<int> s;
map<int, int> mm;
int num = 1, arr[1111111];
vector<int> v[555];
int seg[1111111 * 4], start = 1;
long long int dp[555][555];
int getn(int l, int r) {
  l += start;
  r += start;
  int ans = 1e9;
  while (l <= r) {
    if (l % 2) {
      ans = min(ans, seg[l]);
      l++;
    }
    if ((r % 2) == 0) {
      ans = min(ans, seg[r]);
      r--;
    }
    l /= 2;
    r /= 2;
  }
  return ans;
}
long long int recur(int l, int r, int lx, int rx) {
  if (l >= r) return 1;
  if (dp[lx][rx] != -1) return dp[lx][rx];
  dp[lx][rx] = 0;
  long long int a1 = 0, a2 = 0;
  long long int vv = 1;
  int first = getn(l, r);
  for (int e = 1; e < (int)v[first].size(); e++)
    vv = (vv * recur(v[first][e - 1] + 1, v[first][e] - 1,
                     arr[v[first][e - 1] + 1], arr[v[first][e] - 1])) %
         mod;
  int ll = len[first].first, rr = len[first].second;
  int ll1 = ll, rr1 = rr;
  while (ll >= l) {
    a1 = (a1 + recur(l, ll - 1, arr[l], arr[ll - 1]) *
                   recur(ll, ll1 - 1, arr[ll], arr[ll1 - 1])) %
         mod;
    ll--;
    if (ll < l) break;
    ll = len[arr[ll]].first;
  }
  while (rr <= r) {
    a2 = (a2 + recur(rr1 + 1, rr, arr[rr1 + 1], arr[rr]) *
                   recur(rr + 1, r, arr[rr + 1], arr[r])) %
         mod;
    rr++;
    if (rr > r) break;
    rr = len[arr[rr]].second;
  }
  long long int ans = (a1 * a2) % mod;
  ans = (ans * vv) % mod;
  dp[lx][rx] = ans;
  return dp[lx][rx];
}
int main(void) {
  memset(dp, -1, sizeof(dp));
  for (int e = 0; e < 555; e++) {
    len[e].first = 1e9;
    len[e].second = -1e9;
  }
  int n, m;
  scanf("%d%d", &m, &n);
  for (int e = 0; e < n; e++) {
    scanf("%d", &arr[e]);
    s.insert(arr[e]);
  }
  for (set<int>::iterator it = s.begin(); it != s.end(); it++) mm[*it] = num++;
  for (int e = 0; e < n; e++) arr[e] = mm[arr[e]];
  for (int e = 0; e < n; e++) {
    len[arr[e]].first = min(len[arr[e]].first, e);
    len[arr[e]].second = max(len[arr[e]].second, e);
    v[arr[e]].push_back(e);
  }
  while (start < n) start *= 2;
  for (int e = 0; e < start * 2; e++) seg[e] = 1e9;
  for (int e = 0; e < n; e++) seg[e + start] = arr[e];
  for (int e = start - 1; e >= 1; e--) seg[e] = min(seg[e * 2], seg[e * 2 + 1]);
  for (int e = 1; e < num; e++)
    if (getn(len[e].first, len[e].second) < e) return !printf("0");
  printf("%lld\n", recur(0, n - 1, arr[0], arr[n - 1]));
  return 0;
}