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;
  const double eps = 1e-15, pi = 3.1415926535897932385;
  const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1}, inf = 0x7fffffff,
            inf_ = 0x80000000;
  const long long Inf = 0x7fffffffffffffffLL, Inf_ = 0x8000000000000000LL;
  template <typename T>
  inline void read(T &x) {
    x = 0;
    bool flag = false;
    char ch = getchar();
    while (ch < '0' || ch > '9') flag = ch == '-' ? true : false, ch = getchar();
    while (ch >= '0' && ch <= '9')
      x = (x << 3) + (x << 1) + (ch & 15), ch = getchar();
    if (flag) x = -x;
  }
  int gcd(int x, int y) { return y ? gcd(y, x % y) : x; }
  int lcm(int x, int y) { return x / gcd(x, y) * y; }
  template <typename T>
  void cmx(T &x, T y) {
    if (y > x) x = y;
  }
  template <typename T>
  void cmn(T &x, T y) {
    if (y < x) x = y;
  }
  inline void ad(int &x, int y) {
    x += y;
    if (x >= 998244353) x -= 998244353;
  }
  inline void su(int &x, int y) {
    x -= y;
    if (x < 0) x += 998244353;
  }
  int Pow(int x, int y) {
    int now = 1;
    for (; y; y >>= 1, x = 1LL * x * x % 998244353)
      if (y & 1) now = 1LL * now * x % 998244353;
    return now;
  }
  int a[500005], n, f[500005], u[500005], w[500005], top, sum;
  int main() {
    read(n);
    for (int i = 1; i <= (n); i++) read(a[i]);
    f[0] = 1;
    int op = 1;
    for (int i = (1); i <= (n); i++) {
      int cur = 0;
      while (top > 0 && w[top] >= a[i]) {
        su(sum, 1LL * w[top] * u[top] % 998244353), ad(cur, u[top--]);
      }
      w[++top] = a[i], u[top] = cur;
      if (op)
        ad(u[top], f[i - 1]);
      else
        su(u[top], f[i - 1]);
      ad(sum, 1LL * w[top] * u[top] % 998244353);
      op ^= 1;
      if (op)
        su(f[i], sum);
      else
        ad(f[i], sum);
    }
    printf("%d\n", f[n]);
    return 0;
  }
  