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>
const double PI = acos(-1.0);
using namespace std;
int setb(int n, int pos) { return n = n | (1 << pos); }
int resb(int n, int pos) { return n = n & ~(1 << pos); }
bool checkb(int n, int pos) { return (bool)(n & (1 << pos)); }
inline long long bigmod(long long b, long long p, long long m) {
  long long ret = 1, cur = b % m;
  while (p) {
    if (p & 1) ret = (ret * cur) % m;
    cur = (cur * cur) % m;
    p = p >> 1;
  }
  return ret;
}
int n, m, ara[505];
long long dp[502][502];
bool visit[502][502];
long long F(int l, int r) {
  if (l > r) return 1;
  if (visit[l][r]) return dp[l][r];
  int mn = 1e6, idx;
  for (int i = l; i <= r; i++) {
    if (ara[i] < mn) {
      mn = ara[i];
      idx = i;
    }
  }
  long long ret = 0;
  long long lft = 0;
  for (int i = l; i <= idx; i++)
    lft = (lft + F(l, i - 1) * F(i, idx - 1)) % 998244353;
  long long rgt = 0;
  for (int j = idx; j <= r; j++)
    rgt = (rgt + F(j + 1, r) * F(idx + 1, j)) % 998244353;
  ret = (lft * rgt) % 998244353;
  visit[l][r] = 1;
  return dp[l][r] = ret;
}
int main() {
  int i;
  scanf("%d %d", &n, &m);
  for (i = 1; i <= n; i++) scanf("%d", &ara[i]);
  long long ret = F(1, n);
  printf("%lld", ret);
  printf("\n");
  return 0;
}