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>
#pragma comment(linker, "/STACK:1000000000")
using namespace std;
const int maxn = 510, maxm = (int)1e6 + 10, mod = 998244353;
int n, m;
int v[maxm];
int l[maxn];
int r[maxn];
int dp[maxn][maxn];
int t[4 * maxm];
void build(int it, int l, int r) {
  if (l == r) {
    t[it] = v[l];
  } else {
    int m = (l + r) >> 1;
    build(2 * it, l, m);
    build(2 * it + 1, m + 1, r);
    t[it] = min(t[2 * it], t[2 * it + 1]);
  }
}
int go(int it, int l, int r, int lm, int rm) {
  if (l == lm && r == rm) {
    return t[it];
  } else {
    int m = (l + r) >> 1;
    if (rm <= m) {
      return go(2 * it, l, m, lm, rm);
    } else if (lm > m) {
      return go(2 * it + 1, m + 1, r, lm, rm);
    } else {
      return min(go(2 * it, l, m, lm, m), go(2 * it + 1, m + 1, r, m + 1, rm));
    }
  }
}
vector<int> ps[maxn];
int go(int left, int right) {
  if (left > right) {
    return 1;
  }
  int lc = v[left];
  int rc = v[right];
  if (left != l[lc] || right != r[rc]) {
    return 0;
  }
  if (dp[lc][rc] != -1) {
    return dp[lc][rc];
  }
  int lm = l[lc];
  int rm = r[rc];
  int cur = go(1, 0, m - 1, lm, rm);
  for (int i = cur; i <= n; i++) {
    int li = l[i];
    int ri = r[i];
    if (li > rm) {
      continue;
    }
    if (ri < lm) {
      continue;
    }
    if (!(lm <= li && ri <= rm)) {
      return dp[lc][rc] = 0;
    }
  }
  int s = 0;
  int t = 0;
  int ans = 1;
  for (int i = 0; i < (int)ps[cur].size() - 1; i++) {
    int li = ps[cur][i] + 1;
    int ri = ps[cur][i + 1] - 1;
    if (li <= ri) {
      ans = (long long)ans * go(li, ri) % mod;
    }
  }
  int lcur = l[cur];
  int rcur = r[cur];
  if (ans == 0) {
    return dp[lc][rc] = 0;
  }
  if (lcur == lm) {
    s = 1;
  } else {
    s = go(lm, lcur - 1);
    for (int i = 1; i <= n; i++) {
      int li = l[i];
      int ri = r[i];
      if (li < lcur && li >= lm) {
        s = (s + (long long)go(li, lcur - 1) * go(lm, li - 1)) % mod;
      }
    }
  }
  if (rm == rcur) {
    t = 1;
  } else {
    t = go(rcur + 1, rm);
    for (int i = 1; i <= n; i++) {
      int li = l[i];
      int ri = r[i];
      if (ri > rcur && ri <= rm) {
        t = (t + (long long)go(rcur + 1, ri) * go(ri + 1, rm)) % mod;
      }
    }
  }
  return dp[lc][rc] = (((long long)ans * s) % mod) * t % mod;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    scanf("%d", &v[i]);
  }
  memset(l, -1, sizeof l);
  for (int i = 0; i < m; i++) {
    ps[v[i]].push_back(i);
    if (l[v[i]] == -1) {
      l[v[i]] = i;
    }
    r[v[i]] = i;
  }
  build(1, 0, m - 1);
  memset(dp, -1, sizeof dp);
  cout << go(0, m - 1) << endl;
}