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 a[1000005];
int lpos[505], rpos[505];
int minn[2005][2005], minl[2005][2005], maxr[2005][2005];
bool vis[2005][2005], can[2005][2005];
int f[2005][2005];
int dp(int l, int r) {
  if (l > r) return 1;
  if (!can[l][r]) return 0;
  if (vis[l][r]) return f[l][r];
  vis[l][r] = 1;
  int x = minn[l][r], nl = lpos[x], nr = rpos[x];
  long long s1 = 0, s2 = 0;
  for (int i = nl; i >= l; i--)
    s1 = (s1 + (long long)dp(l, i - 1) * dp(i, nl - 1)) % 998244353;
  for (int i = nr; i <= r; i++)
    s2 = (s2 + (long long)dp(i + 1, r) * dp(nr + 1, i)) % 998244353;
  long long s = s1 * s2 % 998244353;
  for (int i = nl + 1, last2 = nl; i <= nr; i++)
    if (a[i] == x) {
      s = s * dp(last2 + 1, i - 1) % 998244353;
      last2 = i;
    }
  return f[l][r] = s;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  int cnt = 0;
  for (int i = 1; i <= m; i++) {
    int x;
    scanf("%d", &x);
    if (i == 1 || x != a[cnt]) {
      a[++cnt] = x;
      if (!lpos[x]) lpos[x] = cnt;
      rpos[x] = cnt;
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j < i; j++)
      if ((lpos[j] > lpos[i] && lpos[j] < rpos[i]) ||
          (rpos[j] > lpos[i] && rpos[j] < rpos[i])) {
        puts("0");
        return 0;
      }
  if (cnt > 1100) {
    puts("0");
    return 0;
  }
  for (int i = 1; i <= cnt; i++) {
    minn[i][i] = a[i];
    minl[i][i] = lpos[a[i]];
    maxr[i][i] = rpos[a[i]];
    for (int j = i + 1; j <= cnt; j++) {
      minn[i][j] = min(minn[i][j - 1], a[j]);
      minl[i][j] = min(minl[i][j - 1], lpos[a[j]]);
      maxr[i][j] = max(maxr[i][j - 1], rpos[a[j]]);
    }
  }
  for (int i = 1; i <= cnt; i++)
    for (int j = i; j <= cnt; j++)
      can[i][j] = (minl[i][j] >= i && maxr[i][j] <= j);
  printf("%d\n", dp(1, cnt));
  return 0;
}