description: You are given an array of n positive integers a_1, a_2, …, a_n. Your task is to calculate the number of arrays of n positive integers b_1, b_2, …, b_n such that: 
  
    * 1 ≤ b_i ≤ a_i for every i (1 ≤ i ≤ n), and 
    * b_i ≠ b_{i+1} for every i (1 ≤ i ≤ n - 1). 
  
  
  
  The number of such arrays can be very large, so print it modulo 998 244 353.
  
  Input
  
  The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the array a.
  
  The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9).
  
  Output
  
  Print the answer modulo 998 244 353 in a single line.
  
  Examples
  
  Input
  
  
  3
  2 2 2
  
  
  Output
  
  
  2
  
  Input
  
  
  2
  2 3
  
  
  Output
  
  
  4
  
  Input
  
  
  3
  1 1 1
  
  
  Output
  
  
  0
  
  Note
  
  In the first test case possible arrays are [1, 2, 1] and [2, 1, 2].
  
  In the second test case possible arrays are [1, 2], [1, 3], [2, 1] and [2, 3].
solution: #include <bits/stdc++.h>
  using namespace std;
  template <int MOD_>
  struct modnum {
    static constexpr int MOD = MOD_;
    static_assert(MOD_ > 0, "MOD must be positive");
  
   private:
    using ll = long long;
    int v;
    static int minv(int a, int m) {
      a %= m;
      assert(a);
      return a == 1 ? 1 : int(m - ll(minv(m, a)) * ll(m) / a);
    }
  
   public:
    modnum() : v(0) {}
    modnum(ll v_) : v(int(v_ % MOD)) {
      if (v < 0) v += MOD;
    }
    explicit operator int() const { return v; }
    friend std::ostream& operator<<(std::ostream& out, const modnum& n) {
      return out << int(n);
    }
    friend std::istream& operator>>(std::istream& in, modnum& n) {
      ll v_;
      in >> v_;
      n = modnum(v_);
      return in;
    }
    friend bool operator==(const modnum& a, const modnum& b) {
      return a.v == b.v;
    }
    friend bool operator!=(const modnum& a, const modnum& b) {
      return a.v != b.v;
    }
    modnum inv() const {
      modnum res;
      res.v = minv(v, MOD);
      return res;
    }
    friend modnum inv(const modnum& m) { return m.inv(); }
    modnum neg() const {
      modnum res;
      res.v = v ? MOD - v : 0;
      return res;
    }
    friend modnum neg(const modnum& m) { return m.neg(); }
    modnum operator-() const { return neg(); }
    modnum operator+() const { return modnum(*this); }
    modnum& operator++() {
      v++;
      if (v == MOD) v = 0;
      return *this;
    }
    modnum& operator--() {
      if (v == 0) v = MOD;
      v--;
      return *this;
    }
    modnum& operator+=(const modnum& o) {
      v += o.v;
      if (v >= MOD) v -= MOD;
      return *this;
    }
    modnum& operator-=(const modnum& o) {
      v -= o.v;
      if (v < 0) v += MOD;
      return *this;
    }
    modnum& operator*=(const modnum& o) {
      v = int(ll(v) * ll(o.v) % MOD);
      return *this;
    }
    modnum& operator/=(const modnum& o) { return *this *= o.inv(); }
    friend modnum operator++(modnum& a, int) {
      modnum r = a;
      ++a;
      return r;
    }
    friend modnum operator--(modnum& a, int) {
      modnum r = a;
      --a;
      return r;
    }
    friend modnum operator+(const modnum& a, const modnum& b) {
      return modnum(a) += b;
    }
    friend modnum operator-(const modnum& a, const modnum& b) {
      return modnum(a) -= b;
    }
    friend modnum operator*(const modnum& a, const modnum& b) {
      return modnum(a) *= b;
    }
    friend modnum operator/(const modnum& a, const modnum& b) {
      return modnum(a) /= b;
    }
  };
  template <typename T>
  T pow(T a, long long b) {
    assert(b >= 0);
    T r = 1;
    while (b) {
      if (b & 1) r *= a;
      b >>= 1;
      a *= a;
    }
    return r;
  }
  using num = modnum<998244353>;
  vector<num> fact, ifact;
  void init() {
    int N = 1100000;
    fact = {1};
    for (int i = 1; i < N; i++) fact.push_back(i * fact[i - 1]);
    ifact.resize(N);
    ifact.back() = 1 / fact.back();
    for (int i = N - 1; i > 0; i--) ifact[i - 1] = i * ifact[i];
  }
  num ncr(int n, int k) {
    if (k < 0 || k > n) return 0;
    return fact[n] * ifact[k] * ifact[n - k];
  }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(nullptr);
    init();
    int n;
    cin >> n;
    vector<int> r(n);
    for (int& a : r) cin >> a;
    vector<num> dp(n + 1, 0);
    dp[0] = 1;
    vector<num> psum(n + 2, 0);
    psum[1] = 1;
    vector<pair<int, int> > stk;
    stk.push_back({0, -1});
    num Q = 0;
    for (int i = 1; i <= n; i++) {
      while (stk.back().first >= r[i - 1]) {
        num val = stk.back().first;
        int z = stk.back().second;
        stk.pop_back();
        int y = stk.back().second;
        Q -= val * (psum[z + 1] - psum[y + 1]);
      }
      {
        num val = r[i - 1];
        int y = stk.back().second;
        stk.push_back({r[i - 1], i - 1});
        int z = stk.back().second;
        Q += val * (psum[z + 1] - psum[y + 1]);
      }
      dp[i] -= Q;
      psum[i + 1] = psum[i] + dp[i];
    }
    cout << dp[n] * pow(num(-1), n) << '\n';
  }
  