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;
namespace fast_IO {
const int IN_LEN = 10000000, OUT_LEN = 10000000;
char ibuf[IN_LEN], obuf[OUT_LEN], *ih = ibuf + IN_LEN, *oh = obuf,
                                  *lastin = ibuf + IN_LEN,
                                  *lastout = obuf + OUT_LEN - 1;
inline char getchar_() {
  return (ih == lastin) &&
                 (lastin = (ih = ibuf) + fread(ibuf, 1, IN_LEN, stdin),
                  ih == lastin)
             ? EOF
             : *ih++;
}
inline void putchar_(const char x) {
  if (oh == lastout) fwrite(obuf, 1, oh - obuf, stdout), oh = obuf;
  *oh++ = x;
}
inline void flush() { fwrite(obuf, 1, oh - obuf, stdout); }
int read() {
  int x = 0;
  int zf = 1;
  char ch = ' ';
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar_();
  if (ch == '-') zf = -1, ch = getchar_();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar_();
  return x * zf;
}
void write(long long x) {
  if (x < 0) putchar_('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar_(x % 10 + '0');
}
}  // namespace fast_IO
using namespace fast_IO;
const int MOD = 998244353;
long long f[505][505], a[505];
int main() {
  int n = read(), m = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  for (int i = 1; i <= n + 1; ++i) f[i][i] = f[i][i - 1] = 1;
  for (int len = 2; len <= n; ++len) {
    for (int i = 1, j = len; j <= n; ++i, ++j) {
      int pos = min_element(a + i, a + j + 1) - a;
      int val1 = 0, val2 = 0;
      for (int k = i; k <= pos; k++)
        (val1 += f[i][k - 1] * f[k][pos - 1] % MOD) %= MOD;
      for (int k = pos; k <= j; k++)
        (val2 += f[pos + 1][k] * f[k + 1][j] % MOD) %= MOD;
      f[i][j] = (long long)(val1)*val2 % MOD;
    }
  }
  write(f[1][n]);
  flush();
  return 0;
}