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 maxn = 1e6 + 5;
const int mod = 998244353;
int n, m;
int a[maxn];
int nl[1005], nr[1005];
vector<int> vec;
long long f[1005][1005];
struct ex {
  int l, r;
};
ex p[1005];
long long dfs(int l, int r) {
  int i, j;
  int x = 0, y;
  long long L = 0, R = 0;
  if (f[l][r] != -1) return f[l][r];
  f[l][r] = 1;
  if (l > r) return 1;
  for (i = 1; i <= n; i++) {
    if (l <= p[i].l and p[i].r <= r) {
      x = i;
      break;
    }
  }
  if (!x) return 1;
  f[l][r] = dfs(p[x].l + 1, p[x].r - 1);
  L = dfs(l, p[x].l - 1);
  for (i = p[x].l - 1; i >= l;) {
    j = nl[i];
    if (j == 0x3f3f3f3f) {
      i--;
      break;
    }
    L = (L + dfs(j, p[x].l - 1) * dfs(l, j - 1)) % mod;
    i = j - 1;
  }
  R = dfs(p[x].r + 1, r);
  for (i = p[x].r + 1; i <= r;) {
    j = nr[i];
    if (j == 0) {
      i++;
      break;
    }
    R = (R + dfs(p[x].r + 1, j) * dfs(j + 1, r)) % mod;
    i = j + 1;
  }
  f[l][r] = f[l][r] * L % mod * R % mod;
  return f[l][r];
}
int main() {
  int i, j;
  int x, y;
  scanf("%d%d", &n, &m);
  for (i = 1; i <= m; i++) {
    scanf("%d", &x);
    if (!vec.size() or vec.back() != x) {
      vec.push_back(x);
    }
  }
  m = 0;
  for (i = 1; i <= n; i++) {
    p[i].l = 0x3f3f3f3f;
    p[i].r = 0;
  }
  for (auto x : vec) {
    a[++m] = x;
    p[a[m]].l = min(p[a[m]].l, m);
    p[a[m]].r = max(p[a[m]].r, m);
  }
  for (i = n; i > 0; i--) {
    for (j = n; j > i; j--) {
      if (max(p[j].l, p[i].l) <= min(p[j].r, p[i].r) and
          !(p[i].l <= p[j].l and p[j].r <= p[i].r)) {
        printf("0\n");
        return 0;
      } else if (max(p[j].l, p[i].l) <= min(p[j].r, p[i].r)) {
        p[i].l = min(p[i].l, p[j].l);
        p[i].r = max(p[i].r, p[j].r);
      }
    }
  }
  for (i = 1; i <= n; i++) {
    for (j = p[i].l; j <= p[i].r; j++) {
      if (a[j] < i) {
        printf("0\n");
        return 0;
      }
    }
  }
  memset(nl, 0x3f, sizeof(nl));
  for (i = 1; i <= n; i++) {
    nl[p[i].r] = min(nl[p[i].r], p[i].l);
    nr[p[i].l] = max(nr[p[i].l], p[i].r);
  }
  memset(f, -1, sizeof(f));
  dfs(1, m);
  printf("%lld\n", f[1][m]);
  return 0;
}