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>
  const int P = 998244353;
  using namespace std;
  template <class T>
  using vector2D = vector<vector<T>>;
  template <class T>
  using vector3D = vector<vector2D<T>>;
  template <class T>
  using pque = priority_queue<T>;
  template <class T>
  using heap = priority_queue<T, vector<T>, greater<T>>;
  template <class T>
  T max(T a, T b, T c) {
    return max(max(a, b), c);
  }
  template <class T>
  T min(T a, T b, T c) {
    return min(min(a, b), c);
  }
  template <class T>
  bool umax(T &x, T y) {
    return (x < y) ? x = y, true : false;
  }
  template <class T>
  bool umin(T &x, T y) {
    return (x > y) ? x = y, true : false;
  }
  void prepare() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  const int maxn = 200005;
  int top;
  long long mul, add;
  long long num[maxn];
  long long stk[maxn];
  long long sum;
  long long power(long long a, long long b, long long c = P) {
    long long r = 1;
    for (r %= c, a %= c; b; b >>= 1, a = a * a % c)
      if (b & 1) r = r * a % c;
    return r;
  }
  int main() {
    prepare();
    int n;
    long long y;
    cin >> n;
    cin >> y;
    top = 1;
    mul = 1;
    add = 0;
    sum = y;
    stk[1] = y;
    num[1] = 1;
    for (int i = 2; i <= n; i++) {
      long long x;
      cin >> x;
      if (x <= y) {
        long long sum0 = sum;
        while (x < stk[top]) {
          long long l = max(x, stk[top - 1]);
          sum = (sum - (num[top] * mul + add) % P * (stk[top] - l) % P + P) % P;
          --top;
        }
        if (x > stk[top]) {
          ++top;
          stk[top] = x;
        }
        add *= -1;
        mul *= -1;
        sum *= -1;
        add = (add + sum0) % P;
        sum = (sum + sum0 * x) % P;
      } else {
        long long sum0 = sum;
        add *= -1;
        mul *= -1;
        sum *= -1;
        add = (add + sum0) % P;
        sum = (sum + sum0 * x) % P;
        stk[++top] = x;
        num[top] = (sum0 - add) * power(mul, P - 2) % P;
      }
      y = x;
    }
    cout << ((sum + P) % P + P) % P << endl;
    return 0;
  }
  