<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

321. The Spy Network
Time limit per test: 0.5 second(s)
Memory limit: 65536 kilobytes
input: standard
output: standard



The network of spies consists of N intelligence officers. They are numbered with the code numbers from 1 to N so that nobody could discover them. The number 1 belongs to the radiowoman Kat. There is exactly N - 1 communication channels between the spies. It is known that a message from any spy to Kat can reach her. All channels are unidirectional.

A channel can have one of two types: protected and almost protected. It is known that a message will not be intercepted almost surely by the hostile security service if at least half of the channels along the path to radiowoman Kat are protected. What is the minimum number of channels to be made protected from almost protected, so that any message from any spy will not be intercepted almost surely ? What are those channels?

Input
The first line of the input contains the integer number N (1 ≤ N ≤ 200000). The following N - 1 lines contain the description of the communication channels. Each channel is described by a pair of the code numbers of spies (the direction of the channel is from the first spy to the second one) and the parameter pi. If pi = protected, the channel is protected and if pi = almost protected, the channel is almost protected.

Output
Write the number of channels to be converted to protected to the first line of the output. To the next line write numbers of channels to be made protected. If there are several solutions, choose any of them.

Example(s)
sample input
sample output
5
5 1 almost protected
3 1 almost protected
2 3 protected
4 3 almost protected
2
1 2

<|response|>
1. Abridged problem statement
We have N spies (nodes), numbered 1…N, and N−1 one‐way channels forming a directed tree pointing toward spy 1 (the root), so every other spy can reach 1 uniquely. Each channel is either "protected" or "almost protected."
A path is safe if at least half of its edges are protected. We want to upgrade (convert) as few almost‐protected channels to protected as possible so that **every** path from any spy to spy 1 is safe. Output the minimum number of upgrades and any valid list of channel indices to upgrade.

2. Key observations
- Because there are N−1 edges and every node except 1 has exactly one outgoing edge on the unique path to 1, the structure is a rooted tree directed toward 1.
- For any node at depth d (number of edges on its path to root), let k = number of almost-protected edges on that path. Safety requires:
    protected ≥ almost
    ⇒ (d − k) ≥ k
    ⇒ 2k ≤ d
- As we do a DFS from the root (treating the input edges as undirected), we maintain the set S of all almost-protected edges on the current root→u path, keyed by their depths and input indices.
- If at a node u of depth d we ever have 2|S| > d, we must immediately upgrade (remove from S) one almost-protected edge. To maximize its effect on deeper nodes, we should upgrade the edge **closest to the root**, i.e., the one in S with minimum depth.
- This greedy never wastes upgrades and ensures all root→u paths satisfy safety. Each edge is inserted and removed at most once, and each upgrade is one set operation, giving O(N log N) time.

3. Full solution approach
a. Read N, then for each of the N−1 channels read (u, v, type). We number input channels 1…N−1 in reading order.
b. Build an undirected adjacency list. Each entry carries (neighbor, tag), where tag is the input index for an almost-protected edge and −1 for a protected edge.
c. Declare a set `almost_protected` of pairs (depth_of_edge, edge_index) for the currently active almost edges. Also prepare a vector `ans` to collect upgrades.
d. DFS(node u, parent p, integer depth):
   1. If 2*|S| > depth, remove the element of S with smallest `depth_of_edge` (that is `*S.begin()`), record its `edge_index` in `ans`.
   2. For each neighbor v of u (skipping the parent) via edge tag `i`:
      - If `i != -1` (almost edge), insert `(depth, i)` into S.
      - Recursively call DFS(v, u, depth+1).
      - Upon return, if it is an almost edge and `(depth, i)` is still in S, erase it (backtrack).
e. Start DFS from root = 1 (zero‐based: node 0) with depth = 0.
f. Print the size of `ans` and then the list of indices.

4. C++ implementation
```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n;
vector<vector<pair<int, int>>> adj;

void read() {
    cin >> n;
    adj.assign(n, {});
    for(int i = 1; i < n; i++) {
        int u, v;
        cin >> u >> v;
        u--, v--;

        string type;
        getline(cin, type);
        if(type[1] == 'a') {
            adj[u].emplace_back(v, i);
            adj[v].emplace_back(u, i);
        } else if(type[1] == 'p') {
            adj[u].emplace_back(v, -1);
            adj[v].emplace_back(u, -1);
        }
    }
}

void solve() {
    // All channels point towards Kat (node 1), forming a tree rooted there.
    // For a node at depth d (the path to Kat has d edges) we need at least
    // ceil(d / 2) of those edges protected, i.e. at most floor(d / 2) may
    // stay almost protected. We DFS from the root and maintain the set
    // almost_protected of the almost-protected edges on the current
    // root-to-node path, keyed by (depth_of_edge, edge_index).
    //
    // When entering a node at depth d, if the number of almost-protected
    // edges on the path exceeds floor(d / 2), we must upgrade one of them.
    // We greedily upgrade the shallowest such edge (smallest depth), since
    // it is shared by the most descendants and therefore helps the most
    // subtrees at once; that edge is removed from the set and recorded in
    // ans. Backtracking removes each edge we added on the way down.

    vector<int> ans;
    set<pair<int, int>> almost_protected;

    function<void(int, int, int)> dfs = [&](int u, int p, int depth) {
        if(almost_protected.size() * 2 > depth) {
            auto lowest = almost_protected.begin();
            ans.push_back(lowest->second);
            almost_protected.erase(lowest);
        }

        for(auto [v, i]: adj[u]) {
            if(v == p) {
                continue;
            }
            if(i != -1) {
                almost_protected.insert({depth, i});
            }

            dfs(v, u, depth + 1);
            if(i != -1 && almost_protected.count({depth, i})) {
                almost_protected.erase({depth, i});
            }
        }
    };

    dfs(0, -1, 0);

    cout << ans.size() << endl;
    for(int i = 0; i < ans.size(); i++) {
        cout << ans[i] << " \n"[i + 1 == ans.size()];
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

5. Python implementation with detailed comments
```python
import sys
sys.setrecursionlimit(1 << 25)
input = sys.stdin.readline

N = int(input())
# adj[u] = list of (neighbor v, tag); tag is input index for almost, -1 for protected
adj = [[] for _ in range(N)]

for idx in range(1, N):
    parts = input().split()
    u, v = int(parts[0]) - 1, int(parts[1]) - 1
    # parts[2:] is either ["almost","protected"] or ["protected"]
    isAlmost = (parts[2][0] == 'a')
    tag = idx if isAlmost else -1
    adj[u].append((v, tag))
    adj[v].append((u, tag))

# We keep a sorted list of (depth_of_edge, edge_index) for active almost edges
import bisect
active = []
answer = []

def dfs(u, p, depth):
    # Enforce safety: if 2*|active| > depth, upgrade the earliest edge
    if len(active) * 2 > depth:
        d0, e0 = active.pop(0)  # pop the smallest depth_of_edge
        answer.append(e0)
    # Traverse neighbors
    for (v, tag) in adj[u]:
        if v == p:
            continue
        if tag != -1:
            # insert in sorted order by depth then idx
            bisect.insort(active, (depth, tag))
        dfs(v, u, depth+1)
        # backtrack
        if tag != -1:
            key = (depth, tag)
            pos = bisect.bisect_left(active, key)
            if pos < len(active) and active[pos] == key:
                active.pop(pos)

dfs(0, -1, 0)  # start DFS at root=0 with depth=0

# Output result
print(len(answer))
print(*answer)
```

Explanation of main steps:
- We DFS down the tree from node 1, treating the channels as undirected.
- We maintain `active`, the (depth, index) of almost edges on the current path.
- At each node of depth `d`, if `2*len(active) > d`, we remove the earliest almost edge (shallowest).
- Each removal corresponds to an upgrade; we record its input index.
- This guarantees all root→u paths satisfy the condition with a minimum number of upgrades.
