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;
  long long IOS = []() {
    ios::sync_with_stdio(0);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    return 0;
  }();
  template <class T>
  void _W(const T &x) {
    cout << x;
  }
  template <class T>
  void _W(T &x) {
    cout << x;
  }
  template <class T, class U>
  void _W(const pair<T, U> &x) {
    _W(x.first);
    putchar(' ');
    _W(x.second);
  }
  template <class T>
  void _W(const vector<T> &x) {
    for (auto i = x.begin(); i != x.end(); _W(*i++)) cout << ' ';
  }
  void W() {}
  template <class T, class... U>
  void W(const T &head, const U &...tail) {
    _W(head);
    cout << ", ";
    W(tail...);
  }
  using pii = pair<long long, long long>;
  const long long N = 2e6 + 10, mod = 998244353;
  inline long long quick(long long a, long long b) {
    a %= mod;
    long long res = 1;
    while (b) {
      if (b & 1) res = res * a % mod;
      a = a * a % mod;
      b >>= 1;
    }
    return res;
  }
  long long n, tp;
  long long a[N], f[N], mi[N], sf[N], pre[N];
  void solve() {
    scanf("%d", &n);
    for (long long i = 1; i <= n; i++) scanf("%d", &a[i]);
    tp = 1;
    f[0] = 1;
    sf[tp] = 1;
    mi[tp] = a[1];
    pre[tp] = a[1];
    for (long long i = 1; i <= n; i++) {
      long long A = a[i + 1], sn = 0;
      f[i] = -pre[tp];
      while (tp && A <= mi[tp]) sn = (sn + sf[tp]) % mod, tp--;
      mi[++tp] = A;
      sf[tp] = (sn + f[i]) % mod;
      pre[tp] = (pre[tp - 1] + 1ll * A * sf[tp]) % mod;
    }
    long long ans = ((n & 1 ? -f[n] : f[n]) + mod) % mod;
    printf("%d\n", ans);
  }
  signed main() {
    long long tt = 1;
    while (tt--) {
      solve();
    }
  }
  