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 N = 2e5 + 8, inf = 1e9 + 7, mod = 998244353;
  const long long mod2 = (long long)mod * mod;
  int n;
  int a[N];
  struct Node {
    Node *l, *r;
    int len;
    bool reset_, laz;
    long long la2, val;
    Node(int len)
        : l(NULL), r(NULL), len(len), reset_(false), laz(false), la2(0), val(0) {}
    void reset() {
      reset_ = true;
      laz = la2 = val = 0;
    }
    void upd() {
      laz ^= 1;
      la2 = mod - la2;
      val = mod - val;
    }
    void upd(long long v) {
      la2 += v;
      if (la2 >= mod) la2 -= mod;
      (val += v * len) %= mod;
    }
    void down() {
      if (reset_) {
        l->reset();
        r->reset();
        reset_ = false;
      }
      if (laz) {
        l->upd();
        r->upd();
        laz = false;
      }
      if (la2) {
        l->upd(la2);
        r->upd(la2);
        la2 = 0;
      }
    }
  } * root(new Node(inf));
  void upd(Node *u, int l, int r, int a, int b, long long v) {
    if (b < l || r < a) return;
    if (a <= l && r <= b) {
      if (v == -2)
        u->reset();
      else {
        if (~v)
          u->upd(v);
        else
          u->upd();
      }
      return;
    }
    int m = (l + r) >> 1;
    if (!u->l) u->l = new Node(m - l + 1);
    if (!u->r) u->r = new Node(r - l);
    u->down();
    upd(u->l, l, m, a, b, v);
    upd(u->r, m + 1, r, a, b, v);
    u->val = (u->l->val + u->r->val) % mod;
  }
  long long get(Node *u, int l, int r, int a, int b) {
    if (b < l || r < a) return 0;
    if (a <= l && r <= b) return u->val;
    int m = (l + r) >> 1;
    if (!u->l) u->l = new Node(m - l + 1);
    if (!u->r) u->r = new Node(r - l);
    u->down();
    return get(u->l, l, m, a, b) + get(u->r, m + 1, r, a, b);
  }
  long long sum[N];
  int main() {
    cin.tie(NULL)->sync_with_stdio(false);
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    sum[0] = 1;
    for (int i = 1; i <= n; ++i) {
      sum[i] =
          (a[i] * sum[i - 1] - get(root, 1, inf, 1, min(a[i - 1], a[i])) + mod2) %
          mod;
      upd(root, 1, inf, a[i - 1] + 1, inf, -2);
      upd(root, 1, inf, 1, a[i - 1], -1);
      upd(root, 1, inf, 1, a[i], sum[i - 1]);
    }
    cout << sum[n] << '\n';
  }
  