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 long long MOD = 998244353;
  const long long MAXN = 2e5 + 100;
  long long A[MAXN];
  long long f(long long x) { return (x % MOD + MOD) % MOD; }
  struct st {
    vector<pair<long long, long long>> vect;
    long long cant = 0;
    long long acum = 0;
    long long sum = 0;
    void insert(long long x, long long c) {
      vect.push_back({f(x - acum), c});
      cant = cant + c;
      sum = f(sum + x * c);
    }
    void add(long long x) {
      acum = f(acum + x);
      sum = f(sum + x * cant);
    }
    void erase(long long c) {
      while (c) {
        if (vect.back().second <= c) {
          c -= vect.back().second;
          sum = f(sum - (vect.back().first + acum) * vect.back().second % MOD);
          cant = cant - vect.back().second;
          vect.pop_back();
        } else {
          sum = f(sum - (vect.back().first + acum) * c % MOD);
          cant = cant - c;
          vect.back().second -= c;
          break;
        }
      }
    }
  };
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    long long N;
    cin >> N;
    st pp1, pp2;
    for (long long i = 1; i <= N; i++) {
      cin >> A[i];
    }
    for (long long i = 1; i <= N; i++) {
      if (i == 1) {
        pp1.insert(1, A[i]);
        pp2.insert(f(-1), A[i]);
        continue;
      }
      long long tot = 0;
      if ((i + 1) & 1)
        tot = pp1.sum;
      else
        tot = pp2.sum;
      long long tam = pp1.cant;
      if (tam > A[i]) {
        pp1.erase(tam - A[i]);
        pp2.erase(tam - A[i]);
      }
      if (i & 1) {
        pp1.add(tot);
        pp2.add(f(-tot));
      } else {
        pp1.add(f(-tot));
        pp2.add(tot);
      }
      tam = pp1.cant;
      if (tam < A[i]) {
        if (i & 1) {
          pp1.insert(tot, A[i] - tam);
          pp2.insert(f(-tot), A[i] - tam);
        } else {
          pp1.insert(f(-tot), A[i] - tam);
          pp2.insert(tot, A[i] - tam);
        }
      }
    }
    if (N & 1) {
      cout << pp1.sum << '\n';
    } else {
      cout << pp2.sum << '\n';
    }
    return 0;
  }
  