This is the second subtask of problem F. The only differences between this and the first subtask are the constraints on the value of m and the time limit. It is sufficient to solve this subtask in order to hack it, but you need to solve both subtasks in order to hack the first 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 ≤ 10^6) — 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.

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


2 3
1 2 1


Output


1


Input


2 3
2 1 2


Output


0


Input


7 7
4 5 1 6 2 3 7


Output


165


Input


8 17
1 3 2 2 7 8 2 5 5 4 4 4 1 1 6 1 1


Output


20

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>

In the second example, Alice must first paint segment 0 3 with colour 1 and then segment 1 2 with colour 2. 
#include <bits/stdc++.h>
using namespace std;
template <class T>
using min_heap = priority_queue<T, vector<T>, greater<T>>;
namespace input {
template <class T>
void re(complex<T>& x);
template <class T1, class T2>
void re(pair<T1, T2>& p);
template <class T>
void re(vector<T>& a);
template <class T, size_t SZ>
void re(array<T, SZ>& a);
template <class T>
void reA(T A[], int sz);
template <class T>
void re(T& x) {
  cin >> x;
}
void re(double& x) {
  string t;
  re(t);
  x = stod(t);
}
void re(long double& x) {
  string t;
  re(t);
  x = stold(t);
}
template <class Arg, class... Args>
void re(Arg& first, Args&... rest) {
  re(first);
  re(rest...);
}
template <class T1, class T2>
void re(pair<T1, T2>& p) {
  re(p.first, p.second);
}
template <class T>
void re(vector<T>& a) {
  for (int i = 0; i < ((int)a.size()); i++) re(a[i]);
}
template <class T, size_t SZ>
void re(array<T, SZ>& a) {
  for (int i = 0; i < (SZ); i++) re(a[i]);
}
template <class T>
void reA(T A[], int sz) {
  for (int i = 0; i < (sz); i++) re(A[i]);
}
void setupIO(const string& PROB = "") {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  if (PROB.length() != 0) {
    ifstream infile(PROB + ".in");
    if (infile.good()) {
      freopen((PROB + ".in").c_str(), "r", stdin);
      freopen((PROB + ".out").c_str(), "w", stdout);
    }
  }
}
}  // namespace input
using namespace input;
namespace output {
template <class T1, class T2>
void prD(const pair<T1, T2>& x);
template <class T, size_t SZ>
void prD(const array<T, SZ>& x);
template <class T>
void prD(const vector<T>& x);
template <class T>
void prD(const set<T>& x);
template <class T1, class T2>
void prD(const map<T1, T2>& x);
template <class T1, class T2>
void pr(const pair<T1, T2>& x);
template <class T, size_t SZ>
void pr(const array<T, SZ>& x);
template <class T>
void pr(const vector<T>& x);
template <class T>
void pr(const set<T>& x);
template <class T1, class T2>
void pr(const map<T1, T2>& x);
template <class T>
void prD(const T& x) {
  cout << x;
}
template <class Arg, class... Args>
void prD(const Arg& first, const Args&... rest) {
  prD(first);
  prD(rest...);
}
template <class T1, class T2>
void prD(const pair<T1, T2>& x) {
  prD("{", x.first, ", ", x.second, "}");
}
template <class T>
void prDContain(const T& x) {
  prD("{");
  bool fst = 1;
  for (const auto& a : x) prD(!fst ? ", " : "", a), fst = 0;
  prD("}");
}
template <class T, size_t SZ>
void prD(const array<T, SZ>& x) {
  prDContain(x);
}
template <class T>
void prD(const vector<T>& x) {
  prDContain(x);
}
template <class T>
void prD(const set<T>& x) {
  prDContain(x);
}
template <class T1, class T2>
void prD(const map<T1, T2>& x) {
  prDContain(x);
}
void psD() { prD("\n"); }
template <class Arg>
void psD(const Arg& first) {
  prD(first);
  psD();
}
template <class Arg, class... Args>
void psD(const Arg& first, const Args&... rest) {
  prD(first, " ");
  psD(rest...);
}
template <class T>
void pr(const T& x) {
  cout << x;
}
template <class Arg, class... Args>
void pr(const Arg& first, const Args&... rest) {
  pr(first);
  pr(rest...);
}
template <class T1, class T2>
void pr(const pair<T1, T2>& x) {
  pr(x.first, " ", x.second);
}
template <class T>
void prContain(const T& x) {
  bool fst = 1;
  for (const auto& a : x) pr(!fst ? " " : "", a), fst = 0;
}
template <class T, size_t SZ>
void pr(const array<T, SZ>& x) {
  prContain(x);
}
template <class T>
void pr(const vector<T>& x) {
  prContain(x);
}
template <class T>
void pr(const set<T>& x) {
  prContain(x);
}
template <class T1, class T2>
void pr(const map<T1, T2>& x) {
  prContain(x);
}
void ps() { pr("\n"); }
template <class Arg>
void ps(const Arg& first) {
  pr(first);
  ps();
}
template <class Arg, class... Args>
void ps(const Arg& first, const Args&... rest) {
  pr(first, " ");
  ps(rest...);
}
}  // namespace output
using namespace output;
int mod = 998244353;
int A[1000000];
int sLoc[500], eLoc[500];
vector<int> B;
long long dp(int l, int r);
long long leftMemo[1000][500];
long long getLeft(int l, int minVal) {
  if (leftMemo[l][minVal] != -1) return leftMemo[l][minVal];
  long long ans = 0;
  for (int i = l; i < (sLoc[minVal] + 1); i++) {
    ans += dp(l, i - 1) * dp(i, sLoc[minVal] - 1);
    ans %= mod;
  }
  return leftMemo[l][minVal] = ans;
}
long long rightMemo[1000][500];
long long getRight(int r, int minVal) {
  if (rightMemo[r][minVal] != -1) return rightMemo[r][minVal];
  long long ans = 0;
  for (int j = eLoc[minVal]; j < (r + 1); j++) {
    ans += dp(eLoc[minVal] + 1, j) * dp(j + 1, r);
    ans %= mod;
  }
  return rightMemo[r][minVal] = ans;
}
long long memo[1000][1000];
long long dp(int l, int r) {
  int minVal = 1000000010;
  for (int i = l; i < (r + 1); i++) minVal = min(minVal, B[i]);
  if (minVal == 1000000010) return 1;
  if (sLoc[minVal] < l || eLoc[minVal] > r) return memo[l][r] = 0;
  long long ans = getLeft(l, minVal) * getRight(r, minVal);
  ans %= mod;
  int start = -1;
  for (int i = sLoc[minVal]; i < (eLoc[minVal] + 1); i++) {
    if (B[i] == minVal) {
      if (start != -1) {
        ans *= dp(start, i - 1);
        ans %= mod;
        start = -1;
      }
    } else {
      if (start == -1) start = i;
    }
  }
  return memo[l][r] = ans;
}
int main() {
  setupIO();
  int n, m;
  re(n, m);
  reA(A, m);
  for (int i = 0; i < n; i++) sLoc[i] = -1;
  ;
  for (int i = 0; i < (m); i++) {
    A[i]--;
  }
  for (int i = 0; i < (m); i++) {
    if ((int)B.size() > 0 && B.back() == A[i]) continue;
    B.push_back(A[i]);
  }
  for (int i = 0; i < ((int)B.size()); i++) {
    if (sLoc[B[i]] == -1) {
      sLoc[B[i]] = i;
      eLoc[B[i]] = i;
    } else {
      eLoc[B[i]] = i;
    }
  }
  if ((int)B.size() > 1000) {
    ps(0);
    return 0;
  }
  for (int i = 0; i < ((int)B.size()); i++)
    for (int j = 0; j < (500); j++) leftMemo[i][j] = -1;
  ;
  for (int i = 0; i < ((int)B.size()); i++)
    for (int j = 0; j < (500); j++) rightMemo[i][j] = -1;
  ;
  for (int i = 0; i < ((int)B.size()); i++)
    for (int j = 0; j < ((int)B.size()); j++) memo[i][j] = -1;
  ;
  ps(dp(0, (int)B.size() - 1));
  return 0;
}
