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;
char buffer[100000], *p1, *p2;
template <class T>
void read(T& x) {
  char ch = ((p1 == p2 &&
              (p2 = (p1 = buffer) + fread(buffer, 1, 100000, stdin), p1 == p2))
                 ? EOF
                 : *p1++);
  x = 0;
  bool f = 1;
  while (!('0' <= ch && ch <= '9') && ch != '-')
    ch = ((p1 == p2 &&
           (p2 = (p1 = buffer) + fread(buffer, 1, 100000, stdin), p1 == p2))
              ? EOF
              : *p1++);
  if (ch == '-')
    f = 0,
    ch = ((p1 == p2 &&
           (p2 = (p1 = buffer) + fread(buffer, 1, 100000, stdin), p1 == p2))
              ? EOF
              : *p1++);
  while ('0' <= ch && ch <= '9')
    x = (x << 1) + (x << 3) + ch - '0',
    ch = ((p1 == p2 &&
           (p2 = (p1 = buffer) + fread(buffer, 1, 100000, stdin), p1 == p2))
              ? EOF
              : *p1++);
  x = (f) ? x : -x;
}
int n, m;
int a[605];
bool vis[605][605];
int dp[605][605];
int DP(int l, int r) {
  if (l >= r) return 1;
  if (vis[l][r]) return dp[l][r];
  vis[l][r] = 1;
  int& ans = dp[l][r];
  int mi = n + 1, mip = -1;
  for (register int i = l; i <= r; i++)
    if (mi > a[i]) mi = a[i], mip = i;
  int le = 0, ri = 0;
  for (register int i = l - 1; i < mip; i++)
    le += (long long int)DP(l, i) * DP(i + 1, mip - 1) % 998244353,
        le %= 998244353;
  for (register int i = r + 1; i > mip; i--)
    ri += (long long int)DP(i, r) * DP(mip + 1, i - 1) % 998244353,
        ri %= 998244353;
  return ans = (long long int)le * ri % 998244353;
}
int main() {
  read(n), read(m);
  for (register int i = 1; i <= n; i++) read(a[i]);
  printf("%d", DP(1, n));
  return 0;
}