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;
template <typename T>
void maxtt(T& t1, T t2) {
  t1 = max(t1, t2);
}
template <typename T>
void mintt(T& t1, T t2) {
  t1 = min(t1, t2);
}
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
string direc = "URDL";
const long long MOD2 = (long long)998244353 * (long long)998244353;
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "YES" : "NO");
  exit(0);
}
void addmod(int& x, int y, int mod = 998244353) {
  assert(y >= 0);
  x += y;
  if (x >= mod) x -= mod;
  assert(x >= 0 && x < mod);
}
void et(int x = -1) {
  printf("%d\n", x);
  exit(0);
}
long long fastPow(long long x, long long y, int mod = 998244353) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long gcd1(long long x, long long y) { return y ? gcd1(y, x % y) : x; }
const int M = 1035;
int c[M], L[M], R[M], pre[M], pos[1035];
int rc[M], cnt;
int dp[1035][1035], dpm[1035][1035];
inline int dfs(int l, int r) {
  if (r <= l) return 1;
  if (l + 1 == r && c[l] == c[r]) return 1;
  int& tmp = dp[l][r];
  if (tmp != -1) return tmp;
  int minColor = dpm[l][r];
  int lt = L[minColor], rt = R[minColor];
  int wayL = dfs(l, lt - 1), wayR = dfs(rt + 1, r);
  if (lt > l)
    for (int z = lt - 1; z >= l; z--) {
      assert(L[c[z]] <= z);
      z = L[c[z]];
      addmod(wayL, (long long)dfs(l, z - 1) * dfs(z, lt - 1) % 998244353);
    }
  if (rt < r) {
    for (int z = rt + 1; z <= r; z++) {
      assert(R[c[z]] >= z);
      z = R[c[z]];
      addmod(wayR, (long long)dfs(rt + 1, z) * dfs(z + 1, r) % 998244353);
    }
  }
  tmp = 1;
  for (int z = rt; z != lt; z = pre[z]) {
    tmp = (long long)tmp * dfs(pre[z] + 1, z - 1) % 998244353;
  }
  tmp = (long long)tmp * wayL % 998244353 * wayR % 998244353;
  return tmp;
}
void init() {
  memset(dp, -1, sizeof dp);
  for (int(i) = 1; (i) <= (int)(m); (i)++) {
    dpm[i][i] = c[i];
    for (int j = i + 1; j <= m; j++) {
      dpm[i][j] = min(dpm[i][j - 1], c[j]);
    }
  }
}
void fmain(int tid) {
  scanf("%d%d", &n, &m);
  for (int(i) = 1; (i) <= (int)(m); (i)++) {
    scanf("%d", c + i);
    if (cnt == 0 || rc[cnt] != c[i]) rc[++cnt] = c[i];
  }
  m = cnt;
  for (int(i) = 1; (i) <= (int)(m); (i)++) c[i] = rc[i];
  for (int(i) = 1; (i) <= (int)(m); (i)++) {
    if (L[c[i]] == 0) L[c[i]] = i;
    maxtt(R[c[i]], i);
  }
  for (int(i) = 1; (i) <= (int)(m); (i)++) {
    pre[i] = pos[c[i]];
    pos[c[i]] = i;
  }
  for (int(i) = 1; (i) <= (int)(n); (i)++)
    for (int(j) = 1; (j) <= (int)(n); (j)++)
      if (i != j) {
        if (L[i] < L[j] && R[i] < R[j] && L[j] < R[i]) {
          puts("0");
          return;
        }
        if (L[i] < L[j] && R[i] > R[j] && j < i) {
          puts("0");
          return;
        }
      }
  if (m > n + n) {
    puts("0");
    return;
  }
  init();
  printf("%d\n", dfs(1, m));
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain(i);
  }
  return 0;
}