description: Frog Gorf is traveling through Swamp kingdom. Unfortunately, after a poor jump, he fell into a well of n meters depth. Now Gorf is on the bottom of the well and has a long way up.
  
  The surface of the well's walls vary in quality: somewhere they are slippery, but somewhere have convenient ledges. In other words, if Gorf is on x meters below ground level, then in one jump he can go up on any integer distance from 0 to a_x meters inclusive. (Note that Gorf can't jump down, only up).
  
  Unfortunately, Gorf has to take a break after each jump (including jump on 0 meters). And after jumping up to position x meters below ground level, he'll slip exactly b_x meters down while resting.
  
  Calculate the minimum number of jumps Gorf needs to reach ground level.
  
  Input
  
  The first line contains a single integer n (1 ≤ n ≤ 300 000) — the depth of the well.
  
  The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ i), where a_i is the maximum height Gorf can jump from i meters below ground level.
  
  The third line contains n integers b_1, b_2, …, b_n (0 ≤ b_i ≤ n - i), where b_i is the distance Gorf will slip down if he takes a break on i meters below ground level.
  
  Output
  
  If Gorf can't reach ground level, print -1. Otherwise, firstly print integer k — the minimum possible number of jumps.
  
  Then print the sequence d_1,\,d_2, …,\,d_k where d_j is the depth Gorf'll reach after the j-th jump, but before he'll slip down during the break. Ground level is equal to 0.
  
  If there are multiple answers, print any of them.
  
  Examples
  
  Input
  
  
  3
  0 2 2
  1 1 0
  
  
  Output
  
  
  2
  1 0 
  
  
  Input
  
  
  2
  1 1
  1 0
  
  
  Output
  
  
  -1
  
  
  Input
  
  
  10
  0 1 2 3 5 5 6 7 8 5
  9 8 7 1 5 4 3 2 0 0
  
  
  Output
  
  
  3
  9 4 0 
  
  Note
  
  In the first example, Gorf is on the bottom of the well and jump to the height 1 meter below ground level. After that he slip down by meter and stays on height 2 meters below ground level. Now, from here, he can reach ground level in one jump.
  
  In the second example, Gorf can jump to one meter below ground level, but will slip down back to the bottom of the well. That's why he can't reach ground level.
  
  In the third example, Gorf can reach ground level only from the height 5 meters below the ground level. And Gorf can reach this height using a series of jumps 10 ⇒ 9 \dashrightarrow 9 ⇒ 4 \dashrightarrow 5 where ⇒ is the jump and \dashrightarrow is slipping during breaks.
solution: #include <bits/stdc++.h>
  using namespace std;
  int const N = 3e5 + 3;
  int a[N], b[N], cad[N], t[N];
  bool viz[N];
  int dp[N], prv[N];
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0x0);
    cout.tie(0x0);
    int n;
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    for (int i = 1; i <= n; ++i) {
      cin >> b[i];
      cad[i] = i + b[i];
      prv[i + b[i]] = i;
    }
    queue<int> q;
    q.push(n);
    viz[n] = true;
    dp[n] = 0;
    int nextNode = n - 1;
    fill(dp, dp + n, (1 << 30));
    while (q.size()) {
      int x = q.front();
      q.pop();
      if (a[x] >= x) {
        dp[0] = 1 + dp[x];
        t[0] = x;
        while (q.size()) q.pop();
        continue;
      }
      for (; nextNode >= x - a[x]; --nextNode) {
        int to = cad[nextNode];
        if (!viz[to]) {
          dp[to] = 1 + dp[x];
          viz[to] = true;
          t[to] = x;
          q.push(to);
        }
      }
    }
    if (dp[0] >= (1 << 30)) {
      cout << "-1\n";
      exit(0);
    }
    cout << dp[0] << '\n';
    int node = 0;
    vector<int> sol;
    while (t[node]) {
      sol.push_back(prv[node]);
      node = t[node];
    }
    reverse(sol.begin(), sol.end());
    for (auto y : sol) cout << y << ' ';
    return 0;
  }
  