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 N = 1e3 + 7, M = 1e6 + 7;
const long long MOD = 998244353;
int b[M], c[M], mx[N], mi[N];
long long f[N][N];
long long solve(int l, int r) {
  if (f[l][r] != -1) return f[l][r];
  if (l >= r) return 1;
  int mii = 0x3f3f3f3f, y = -1;
  for (int i = l; i <= r; i++) mii = min(mii, b[i]);
  for (int i = l; i <= r; i++)
    if (b[i] == mii) {
      y = i;
      break;
    }
  long long suml = 0, sumr = 0, ans;
  int mil = 0x3f3f3f3f, mxr = -0x3f3f3f3f;
  for (int i = y; i >= l; i--) {
    mil = min(mil, mi[b[i]]);
    if (mil >= i) suml = (suml + solve(l, i - 1) * solve(i, y - 1) % MOD) % MOD;
  }
  ans = suml;
  for (int i = y + 1; i <= r; i++) {
    if (b[i] == mii) {
      ans = ans * solve(y + 1, i - 1) % MOD;
      y = i;
    }
  }
  for (int i = y; i <= r; i++) {
    mxr = max(mxr, mx[b[i]]);
    if (mxr <= i) sumr = (sumr + solve(y + 1, i) * solve(i + 1, r) % MOD) % MOD;
  }
  return f[l][r] = ans * sumr % MOD;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) scanf("%d", &c[i]);
  int cnt = 0;
  c[0] = -1;
  for (int i = 1; i <= m; i++)
    if (c[i] != c[i - 1]) b[++cnt] = c[i];
  for (int i = 1; i <= n; i++) {
    int j = 1;
    while (b[j] != i) j++;
    int k = cnt;
    while (b[k] != i) k--;
    for (int l = j + 1; l <= k - 1; l++) {
      if (b[l] < i) {
        printf("0");
        return 0;
      }
    }
  }
  memset(mi, 0x3f3f3f3f, sizeof(mi));
  memset(mx, -1, sizeof(mx));
  for (int i = 1; i <= cnt; i++)
    mi[b[i]] = min(mi[b[i]], i), mx[b[i]] = max(mx[b[i]], i);
  memset(f, -1, sizeof(f));
  printf("%lld", solve(1, cnt));
  return 0;
}