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;
using LL = long long;
namespace _buff {
const size_t BUFF = 1 << 19;
char ibuf[BUFF], *ib = ibuf, *ie = ibuf;
char getc() {
  if (ib == ie) {
    ib = ibuf;
    ie = ibuf + fread(ibuf, 1, BUFF, stdin);
  }
  return ib == ie ? -1 : *ib++;
}
}  // namespace _buff
LL read() {
  using namespace _buff;
  LL ret = 0;
  bool pos = true;
  char c = getc();
  for (; (c < '0' || c > '9') && c != '-'; c = getc()) {
    assert(~c);
  }
  if (c == '-') {
    pos = false;
    c = getc();
  }
  for (; c >= '0' && c <= '9'; c = getc()) {
    ret = (ret << 3) + (ret << 1) + (c ^ 48);
  }
  return pos ? ret : -ret;
}
const size_t N = 500 + 5;
const int MOD = 998244353;
void ladd(int &dp, LL val) { dp = (dp + val) % MOD; }
int n, m, c[N], dp[N][N];
int solve(int l, int r) {
  if (l >= r) return 1;
  if (dp[l][r]) {
    return dp[l][r] - 1;
  }
  int pos = l;
  for (int i = l + 1; i <= r; ++i) {
    if (c[i] < c[pos]) {
      pos = i;
    }
  }
  int lft = 0;
  for (int lp = l; lp <= pos; ++lp) {
    ladd(lft, (LL)solve(l, lp - 1) * solve(lp, pos - 1));
  }
  int rht = 0;
  for (int rp = pos; rp <= r; ++rp) {
    ladd(rht, (LL)solve(pos + 1, rp) * solve(rp + 1, r));
  }
  return (dp[l][r] = (LL)lft * rht % MOD + 1) - 1;
}
int main() {
  n = read();
  m = read();
  generate_n(c + 1, m, read);
  cout << solve(1, m) << '\n';
  return 0;
}