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 int md = 998244353;
  struct lzT {
    long long add;
    int mul;
  };
  struct Node {
    long long res = 0;
    lzT lz = {0, 1};
    int hi, lo;
    Node *L, *R;
    Node(int _lo, int _hi) : lo(_lo), hi(_hi) { L = R = nullptr; }
    void extend() {
      if (L or R or lo == hi) return;
      int mid = (lo + hi) / 2;
      L = new Node(lo, mid);
      R = new Node(mid + 1, hi);
    }
    lzT combLz(lzT a, lzT b) {
      return {(a.mul * b.add + a.add + md) % md, a.mul * b.mul};
    }
    long long applyLz() {
      return (res * lz.mul + lz.add * (hi - lo + 1) + md) % md;
    }
    void push() {
      res = applyLz();
      if (lo != hi) {
        L->lz = combLz(lz, L->lz);
        R->lz = combLz(lz, R->lz);
      }
      lz = {0, 1};
    }
    void upd(int ql, int qr, lzT val) {
      if (lo > qr or hi < ql) return;
      if (lo >= ql and hi <= qr) {
        lz = combLz(val, lz);
        return;
      }
      extend();
      push();
      L->upd(ql, qr, val);
      R->upd(ql, qr, val);
      res = (L->applyLz() + R->applyLz()) % md;
    }
    long long qry(int ql, int qr) {
      if (lo > qr or hi < ql) return {};
      if (lo >= ql and hi <= qr) return applyLz();
      extend();
      push();
      return (L->qry(ql, qr) + R->qry(ql, qr)) % md;
    }
  };
  Node* seg = new Node(0, 1e9);
  int main() {
    int n;
    cin >> n;
    for (int i = 0, num; i < n; i++) {
      cin >> num;
      if (!i)
        seg->upd(1, num, {1, 1});
      else {
        long long sm = seg->qry(1, 1e9);
        seg->upd(1, num, {sm, -1});
        seg->upd(num + 1, 1e9, {0, 0});
      }
    }
    cout << seg->qry(1, 1e9) << endl;
  }
  