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 long long MOD = 998244353;
const int base = 1048576;
int T[4 * 1010101];
int L[505];
int R[505];
long long D[505][505];
int C[1010101];
int N, M;
int bef[505];
vector<int> loc[505];
void maketree() {
  for (int i = 1; i <= M; i++) T[base + i - 1] = C[i];
  for (int i = base - 1; i >= 1; i--) T[i] = min(T[i * 2], T[i * 2 + 1]);
}
int query(int s, int e) {
  int l = base + s, r = base + e;
  l--;
  r--;
  int ret = N + 1;
  while (l <= r) {
    if (l & 1) {
      ret = min(ret, T[l]);
      l++;
    }
    if (!(r & 1)) {
      ret = min(ret, T[r]);
      r--;
    }
    l >>= 1;
    r >>= 1;
  }
  return ret;
}
long long f(int l, int r) {
  if (R[r] == L[l]) return 1;
  if (L[l] > R[r]) return 1;
  if (D[l][r] != -1) return D[l][r];
  int m = query(L[l], R[r]);
  long long t = 1;
  for (int i = 1; i < loc[m].size(); i++) {
    if (loc[m][i] - loc[m][i - 1] == 1) continue;
    t = t * f(C[loc[m][i - 1] + 1], C[loc[m][i] - 1]) % MOD;
  }
  long long sl = 0, sr = 0;
  int lidx = loc[m][0] - 1;
  if (lidx < L[l])
    sl = 1;
  else {
    while (1) {
      if (lidx >= L[l])
        sl = (sl + f(C[lidx + 1], C[loc[m][0] - 1]) * f(l, C[lidx])) % MOD;
      else
        sl = (sl + f(C[lidx + 1], C[loc[m][0] - 1])) % MOD;
      if (lidx < L[l]) break;
      lidx = L[C[lidx]] - 1;
    }
  }
  int ridx = loc[m].back() + 1;
  if (ridx > R[r])
    sr = 1;
  else {
    while (1) {
      if (ridx <= R[r])
        sr = (sr + f(C[loc[m].back() + 1], C[ridx - 1]) * f(C[ridx], r)) % MOD;
      else
        sr = (sr + f(C[loc[m].back() + 1], C[ridx - 1])) % MOD;
      if (ridx > R[r]) break;
      ridx = R[C[ridx]] + 1;
    }
  }
  return D[l][r] = t * sl % MOD * sr % MOD;
}
int main() {
  scanf("%d%d", &N, &M);
  for (int i = 1; i <= N; i++) {
    L[i] = M + 1;
    R[i] = 0;
  }
  C[M + 1] = N + 1;
  L[N + 1] = R[N + 1] = N + 1;
  for (int i = 1; i <= M; i++) {
    scanf("%d", &C[i]);
    L[C[i]] = min(L[C[i]], i);
    R[C[i]] = max(R[C[i]], i);
    loc[C[i]].push_back(i);
  }
  maketree();
  for (int i = 1; i <= M; i++) {
    if (bef[C[i]] && query(bef[C[i]] + 1, i - 1) < C[i]) {
      puts("0");
      return 0;
    }
    bef[C[i]] = i;
  }
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= N; j++) {
      D[i][j] = -1;
    }
  }
  printf("%lld\n", f(C[1], C[M]));
  return 0;
}