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 N = 2e5 + 10;
  const long long mod = 998244353;
  const long long MAXL = 60;
  long long n;
  long long a[N], dp[N];
  struct ty {
    long long f, sum;
    long long lz;
  } tr[N << 2];
  vector<long long> st;
  void pushup(long long p, long long l, long long r) {
    tr[p].sum = (tr[p << 1].sum + tr[p << 1 | 1].sum) % mod;
    tr[p].f = (tr[p << 1].f + tr[p << 1 | 1].f) % mod;
  }
  void pushdown(long long p, long long l, long long r) {
    long long x = tr[p].lz;
    tr[p].lz = 0;
    tr[p << 1].lz = tr[p << 1 | 1].lz = x;
    tr[p << 1].f = tr[p << 1].sum * x % mod;
    tr[p << 1 | 1].f = tr[p << 1 | 1].sum * x % mod;
  }
  void insert(long long p, long long l, long long r, long long pos,
              long long val) {
    if (l == r) {
      tr[p].f = val * tr[p].lz % mod;
      tr[p].sum = val;
      return;
    }
    if (tr[p].lz) pushdown(p, l, r);
    long long mid = (l + r) >> 1;
    if (pos <= mid)
      insert(p << 1, l, mid, pos, val);
    else
      insert(p << 1 | 1, mid + 1, r, pos, val);
    pushup(p, l, r);
  }
  void change(long long p, long long l, long long r, long long x, long long y,
              long long val) {
    if (x <= l && r <= y) {
      tr[p].lz = val;
      tr[p].f = tr[p].sum * val % mod;
      return;
    }
    if (tr[p].lz) pushdown(p, l, r);
    long long mid = (l + r) >> 1;
    if (y <= mid)
      change(p << 1, l, mid, x, y, val);
    else if (x >= mid + 1)
      change(p << 1 | 1, mid + 1, r, x, y, val);
    else {
      change(p << 1, l, mid, x, mid, val),
          change(p << 1 | 1, mid + 1, r, mid + 1, y, val);
    }
    pushup(p, l, r);
  }
  void de() {
    cout << " de " << endl;
    for (long long i = (1); i <= (n); i++) cout << dp[i] << " ";
    cout << endl;
  }
  signed main() {
    ios::sync_with_stdio(false), cin.tie(0);
    cin >> n;
    for (long long i = (1); i <= (n); i++) cin >> a[i];
    insert(1, 1, n, 1, 1);
    for (long long i = (1); i <= (n); i++) {
      while (!st.empty() && a[i] < a[st.back()]) st.pop_back();
      long long x = (st.empty()) ? 1 : st.back() + 1;
      st.push_back(i);
      change(1, 1, n, x, i, a[i]);
      dp[i] = ((i & 1) ? tr[1].f : (mod - tr[1].f) % mod);
      if (i < n) insert(1, 1, n, i + 1, (i & 1) ? (mod - dp[i]) % mod : dp[i]);
    }
    cout << dp[n] << endl;
    return 0;
  }
  