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;
inline long long read() {
  long long f = 1, x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + (s - '0');
    s = getchar();
  }
  return x * f;
}
const int jzm = 998244353;
inline int ABS(int x) { return x < 0 ? -x : x; }
int n, m, i, j, s, o, k;
int a[505];
int dp[505][505];
int pos[505][505];
int DP(int l, int r) {
  if (l > r) return 1;
  if (dp[l][r] != -1) return dp[l][r];
  int p = pos[l][r];
  int ls = 0, rs = 0;
  for (int i = l; i <= p; i++) {
    ls = (ls + 0ll + DP(l, i - 1) * 1ll * DP(i, p - 1) % jzm) % jzm;
  }
  for (int i = p; i <= r; i++) {
    rs = (rs + 0ll + DP(p + 1, i) * 1ll * DP(i + 1, r) % jzm) % jzm;
  }
  dp[l][r] = ls * 1ll * rs % jzm;
  return dp[l][r];
}
int main() {
  memset(dp, -1, sizeof(dp));
  n = read();
  m = read();
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    pos[i][i] = i;
    int minn = a[i];
    for (int j = i - 1; j > 0; j--) {
      if (a[j] < minn) {
        minn = a[j];
        pos[j][i] = j;
      } else
        pos[j][i] = pos[j + 1][i];
    }
  }
  printf("%d\n", DP(1, n));
  return 0;
}