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;
vector<int> v[510], v1[510];
long long f[510][510];
int ch[510][2], size[510], tot, root;
int num[1000010], n, m;
inline int rd() {
  int x = 0;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    ;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  return x;
}
inline long long pls(const long long &x, const long long &y) {
  return (x + y < 998244353LL) ? x + y : x + y - 998244353LL;
}
inline long long mns(const long long &x, const long long &y) {
  return (x - y < 0) ? x - y + 998244353LL : x - y;
}
inline long long ksm(long long x, long long y) {
  long long res = 1;
  for (; y; y >>= 1, x = x * x % 998244353LL)
    if (y & 1) res = res * x % 998244353LL;
  return res;
}
int mn[4000010];
inline void build(int o, int l, int r) {
  if (l == r) {
    mn[o] = num[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(o << 1, l, mid);
  build(o << 1 | 1, mid + 1, r);
  mn[o] = min(mn[o << 1], mn[o << 1 | 1]);
}
inline int query(int o, int l, int r, int x, int y) {
  if (l >= x && r <= y) return mn[o];
  int mid = (l + r) >> 1, res = m + 1;
  if (x <= mid) res = min(res, query(o << 1, l, mid, x, y));
  if (y > mid) res = min(res, query(o << 1 | 1, mid + 1, r, x, y));
  return res;
}
inline int build(int l, int r) {
  int now = ++tot, id = query(1, 1, n, l, r);
  size[now] = 1;
  if (v[id][0] < l || v[id][v[id].size() - 1] > r) {
    puts("0");
    exit(0);
  }
  for (int i = 0; i < v[id].size() - 1; i++)
    if (v[id][i] < v[id][i + 1] - 1)
      v1[now].push_back(build(v[id][i] + 1, v[id][i + 1] - 1));
  for (int t : v1[now]) size[now] += size[t];
  if (v[id][0] > l)
    ch[now][0] = build(l, v[id][0] - 1), size[now] += size[ch[now][0]];
  if (v[id][v[id].size() - 1] < r)
    ch[now][1] = build(v[id][v[id].size() - 1] + 1, r),
    size[now] += size[ch[now][1]];
  return now;
}
inline void dfs(int x) {
  if (!x) return;
  if (size[x] == 1) {
    for (int i = 0; i <= m; i++) f[x][i] = i + 1;
    return;
  }
  for (int t : v1[x]) dfs(t);
  dfs(ch[x][0]);
  dfs(ch[x][1]);
  long long res = 1;
  for (int t : v1[x]) res = res * f[t][0] % 998244353LL;
  for (int i = 0; i <= m; i++) {
    for (int j = 0; j <= i; j++)
      f[x][i] = pls(f[x][i], f[ch[x][0]][j + 1] * f[ch[x][1]][i - j + 1] %
                                 998244353LL * res % 998244353LL);
  }
}
int main() {
  m = rd();
  n = rd();
  for (int i = 1; i <= n; i++) v[num[i] = rd()].push_back(i);
  build(1, 1, n);
  root = build(1, n);
  for (int i = 0; i <= m + 1; i++) f[0][i] = 1;
  dfs(root);
  printf("%lld\n", f[root][0]);
  return 0;
}