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;
char buf[1 << 21], *p1 = buf, *p2 = buf;
long long read() {
  long long w = 0, f = 1;
  char c = ' ';
  while (c < '0' || c > '9')
    c = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
             ? EOF
             : *p1++),
    f = c == '-' ? -1 : f;
  while (c >= '0' && c <= '9')
    w = w * 10 + c - 48,
    c = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
             ? EOF
             : *p1++);
  return w * f;
}
long long n, m, len, a[1005], dp[1005][1005], b[1005], l[1005], r[1005];
const long long mod = 998244353;
signed main() {
  n = read(), m = read();
  for (long long i = 1; i <= m; i++) {
    long long x = read();
    if (x != a[len]) a[++len] = x;
  }
  m = len;
  if (m > n << 1) {
    puts("0");
    return 0;
  }
  for (long long i = 1; i <= n; i++) {
    bool flag = 0;
    memset(b, 0, sizeof(b));
    for (long long j = 1; j < i; j++)
      if (a[j] > a[i]) b[a[j]] = 1;
    for (long long j = n; j > i; j--)
      if (a[j] > a[i] && b[a[j]]) {
        flag = 1;
        break;
      }
    if (flag) {
      puts("0");
      return 0;
    }
  }
  for (long long i = 1; i <= n; i++) l[i] = m + 1;
  for (long long i = 1; i <= m; i++)
    l[a[i]] = min(l[a[i]], i), r[a[i]] = max(r[a[i]], i);
  for (long long i = 0; i <= m; i++) dp[i + 1][i] = 1;
  for (long long i = 1; i <= m; i++)
    if (l[a[i]] == r[a[i]] && l[a[i]] == i) dp[i][i] = 1;
  for (long long i = 2; i <= m; i++)
    for (long long j = 1; j <= m - i + 1; j++) {
      long long minn = 1e9, minid, s1 = 0, s2 = 0;
      for (long long k = j; k <= j + i - 1; k++)
        if (a[k] < minn) minn = a[k], minid = k;
      if (l[minn] < j || r[minn] > j + i - 1) continue;
      for (long long k = j; k <= l[minn]; k++)
        s1 = (s1 + dp[j][k - 1] * dp[k][l[minn] - 1]) % mod;
      for (long long k = r[minn]; k <= j + i - 1; k++)
        s2 = (s2 + dp[r[minn] + 1][k] * dp[k + 1][j + i - 1]) % mod;
      dp[j][j + i - 1] = s1 * s2 % mod;
      for (long long k = minid + 1, la = minid; k <= j + i - 1; k++)
        if (a[k] == a[minid])
          dp[j][j + i - 1] = dp[j][j + i - 1] * dp[la + 1][k - 1] % mod, la = k;
    }
  printf("%lld", dp[1][m]);
  return 0;
}