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;
int a[1000005], b[1000005], L[1000005], R[1000005];
int mod = 998244353;
struct Seg {
  struct Tree {
    int l, r, ans;
  } t[4000005];
  void build(int id, int l, int r) {
    t[id].l = l, t[id].r = r;
    if (l == r) {
      t[id].ans = a[l];
      return;
    }
    int mid = (l + r) / 2;
    build(id << 1, l, mid);
    build(id << 1 | 1, mid + 1, r);
    t[id].ans = min(t[id << 1].ans, t[id << 1 | 1].ans);
  }
  int get_ans(int id, int l, int r) {
    if (l <= t[id].l && t[id].r <= r) {
      return t[id].ans;
    }
    int mid = (t[id].l + t[id].r) / 2, tot = 1e9;
    if (l <= mid) tot = min(tot, get_ans(id << 1, l, r));
    if (r > mid) tot = min(tot, get_ans(id << 1 | 1, l, r));
    return tot;
  }
} A;
int n, m;
int lst[1000005], P[1000005];
int dp[1005][1005];
int get_ans(int l, int r) {
  if (l >= r) return 1;
  if (dp[l][r]) return dp[l][r];
  int num = A.get_ans(1, l, r);
  int Ans = 0, Ans1 = 0, Ans2 = 0;
  for (int i = L[num]; i >= l; i--) {
    Ans1 = (1ll * Ans1 +
            (1ll * get_ans(i, L[num] - 1) * get_ans(l, i - 1)) % mod) %
           mod;
  }
  for (int i = R[num]; i <= r; i++) {
    Ans2 = (1ll * Ans2 +
            (1ll * get_ans(R[num] + 1, i) * get_ans(i + 1, r)) % mod) %
           mod;
  }
  Ans = (1ll * Ans1 * Ans2) % mod;
  int now = R[num], pre = lst[now];
  while (1) {
    if (L[num] == R[num]) break;
    Ans = (1ll * Ans * get_ans(pre + 1, now - 1)) % mod;
    now = pre, pre = lst[pre];
    if (now == L[num]) break;
  }
  return dp[l][r] = Ans;
}
signed main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++) scanf("%d", &a[i]);
  int cnt = 0;
  for (int i = 1; i <= m; i++)
    if (a[i] != a[i - 1]) b[++cnt] = a[i];
  for (int i = 1; i <= cnt; i++) a[i] = b[i];
  m = cnt;
  memset(L, 0x3f, sizeof L);
  for (int i = 1; i <= m; i++) {
    lst[i] = P[a[i]];
    L[a[i]] = min(L[a[i]], i);
    R[a[i]] = max(R[a[i]], i);
    P[a[i]] = i;
  }
  A.build(1, 1, m);
  for (int i = 1; i <= n; i++)
    if (P[i] == 0 || A.get_ans(1, L[i], R[i]) != i) return printf("0") & 0;
  printf("%d", get_ans(1, m));
  return 0;
}