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 <typename T>
  using matrix = vector<vector<T>>;
  const long long INFL = (1LL << 62) - 1;
  const int INF = (1 << 30) - 1;
  const double EPS = 1e-7;
  const int MOD = 998244353;
  const int RANDOM =
      chrono::high_resolution_clock::now().time_since_epoch().count();
  const int MAXN = 1e6 + 1;
  struct seg {
    int sum, mult, add, e, d;
  };
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    cin >> n;
    vector<seg> tree;
    function<int()> create = [&] {
      int ret = tree.size();
      tree.push_back({0, 1, 0, 0, 0});
      return ret;
    };
    create();
    function<void(long long, long long, long long)> refresh =
        [&](long long id, long long l, long long r) {
          if (tree[id].mult == 1 && tree[id].add == 0) return;
          int e = tree[id].e, d = tree[id].d;
          if (l != r) {
            tree[e].mult *= tree[id].mult;
            tree[e].add *= tree[id].mult;
            tree[e].add = ((long long)tree[e].add + tree[id].add + MOD) % MOD;
            tree[d].mult *= tree[id].mult;
            tree[d].add *= tree[id].mult;
            tree[d].add = ((long long)tree[d].add + tree[id].add + MOD) % MOD;
          }
          tree[id].sum *= tree[id].mult;
          tree[id].sum =
              (tree[id].sum + (long long)(r - l + 1) * tree[id].add + MOD) % MOD;
          tree[id].mult = 1;
          tree[id].add = 0;
        };
    function<void(long long, long long, long long, long long, long long,
                  long long, long long)>
        update = [&](long long id, long long l, long long r, long long ql,
                     long long qr, long long t, long long val) {
          int e = tree[id].e, d = tree[id].d, m = (l + r) >> 1;
          if (l != r && e == 0) {
            e = create();
            d = create();
            tree[id].e = e, tree[id].d = d;
          }
          refresh(id, l, r);
          if (l > qr || r < ql) return;
          if (ql <= l && r <= qr) {
            if (t) {
              tree[id].mult *= val;
              tree[id].add *= val;
            } else {
              tree[id].add += val;
            }
            refresh(id, l, r);
            return;
          }
          update(e, l, m, ql, qr, t, val);
          update(d, m + 1, r, ql, qr, t, val);
          tree[id].sum = ((long long)tree[e].sum + tree[d].sum + MOD) % MOD;
        };
    long long last = 1;
    for (int i = 0; i < n; i++) {
      int in;
      cin >> in;
      if (in != 1e9) update(0, 1, 1e9, in + 1, 1e9, 1, 0);
      update(0, 1, 1e9, 1, 1e9, 1, -1);
      update(0, 1, 1e9, 1, in, 0, last);
      last = tree[0].sum;
    }
    cout << last << '\n';
    return 0;
  }
  