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;
void read(long long &x) {
  x = 0;
  long long f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -f;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  x *= f;
  return;
}
long long fs(long long x) { return x < 0 ? -x : x; }
void print(long long x) {
  if (x < 10) {
    putchar(x + '0');
    return;
  }
  print(x / 10);
  putchar(x % 10 + '0');
  return;
}
void write(long long x) {
  if (x < 0) {
    putchar('-');
    x = fs(x);
  }
  print(x);
  return;
}
long long mint(long long x, long long y) { return x < y ? x : y; }
long long maxt(long long x, long long y) { return x > y ? x : y; }
long long n, a[3005], dp[3005][3005], g[3005][3005], m;
int main() {
  read(n), read(m);
  for (int i = 1; i <= n; i++) read(a[i]);
  for (int i = 1; i <= n + 1; i++)
    g[i][i] = dp[i][i] = dp[i][i - 1] = g[i][i - 1] = 1ll;
  a[0] = 0x3f3f3f3f3f3f3f;
  for (long long len = 2; len <= n; len++)
    for (long long l = 1; l + len - 1 <= n; l++) {
      long long r = l + len - 1;
      long long p = 0;
      for (long long k = l; k <= r; k++)
        if (a[p] > a[k]) p = k;
      dp[l][r] = g[l][r] = dp[l][p - 1] * dp[p + 1][r] % 998244353;
      for (long long k = l; k < r; k++)
        dp[l][r] = (g[l][k] * dp[k + 1][r] % 998244353 + dp[l][r]) % 998244353;
    }
  printf("%lld\n", dp[1][n]);
}