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;
const int mod = 998244353, INF = 1e9;
int n, m, a[1000002], s[2002], len, l[502], r[502], nex[1000002], num[1000002],
    xyg[1000002];
bool fl[1000002];
long long f[2002][2002];
bool pd(int x, int y) {
  for (int i = x; i <= y; i++)
    if (l[a[s[i]]] < s[x] || r[a[s[i]]] > s[y]) return false;
  return true;
}
long long dp(int x, int y) {
  if (!x || !y) return 0;
  if (y < x) return 1;
  if (l[a[s[y]]] == s[y] && r[a[s[y]]] > l[a[s[y]]] ||
      r[a[s[x]]] == s[x] && r[a[s[x]]] > l[a[s[x]]])
    return 0;
  if (f[x][y] != -1) return f[x][y];
  if (!pd(x, y)) return f[x][y] = 0;
  f[x][y] = 0;
  int Min = INF;
  for (int i = x; i <= y; i++) Min = min(Min, a[s[i]]);
  long long qt = 1;
  for (int i = l[Min]; nex[xyg[i]]; i = nex[i]) {
    i = xyg[i];
    qt = qt * dp(num[i + 1], num[nex[i] - 1]) % mod;
    if (!qt) break;
  }
  if (!qt) return f[x][y] = 0;
  f[x][y] = qt;
  long long s1 = 0, s2 = 0, t;
  for (int i = x; i <= num[l[Min]]; i++) {
    t = 1;
    if (i > x) t = t * dp(x, num[s[i] - 1]) % mod;
    if (i < num[l[Min]]) t = t * dp(num[s[i]], num[l[Min] - 1]) % mod;
    s1 = (s1 + t) % mod;
  }
  for (int i = num[r[Min]]; i <= y; i++) {
    t = 1;
    if (i > num[r[Min]]) t = t * dp(num[r[Min] + 1], num[s[i]]) % mod;
    if (i < y) t = t * dp(num[s[i] + 1], y) % mod;
    s2 = (s2 + t) % mod;
  }
  f[x][y] = f[x][y] * s1 % mod * s2 % mod;
  return f[x][y];
}
int main() {
  memset(f, -1, sizeof(f));
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) scanf("%d", &a[i]);
  xyg[m] = m;
  for (int i = m - 1; i >= 1; i--)
    if (a[i] == a[i + 1])
      xyg[i] = xyg[i + 1];
    else
      xyg[i] = i;
  for (int i = 1; i <= m; i++) {
    nex[r[a[i]]] = i;
    r[a[i]] = i;
  }
  for (int i = m; i >= 1; i--) l[a[i]] = i;
  for (int i = 1; i <= n; i++)
    fl[l[i]] = fl[r[i]] = fl[l[i] - 1] = fl[r[i] + 1] = 1;
  for (int i = 1; i <= m; i++)
    if (fl[i]) {
      s[++len] = i;
      num[i] = len;
    }
  printf("%lld\n", dp(1, len));
  return 0;
}