1. Abridged problem statement  
Given an initial sequence [1, 2, …, N], perform M operations of the form “reverse the subarray from positions l to r” and output the final sequence. Constraints: N≤130 000, M≤2000.

2. Detailed editorial  
We need a data structure that supports reversing arbitrary contiguous subranges in a sequence of size up to 130 000, M up to 2000, and finally outputting the sequence. Doing each reverse in O(r−l+1) time on an array could be as bad as O(N × M)=2.6×10^8 element moves, which may be too slow in low-level languages.

An implicit treap (a randomized balanced binary search tree keyed by position) with lazy propagation of “reverse” flags handles this in O((N + M) log N) time:

1. Representation  
   - Each node stores:  
     · its value (1…N),  
     · its subtree size,  
     · a random priority,  
     · boolean rev flag,  
     · pointers left, right.  

2. Invariants  
   - In-order traversal yields the sequence in current order.  
   - The tree is heap-ordered by priority (max-heap or min-heap).  
   - The implicit “key” is the in-order index; we never store it explicitly.

3. Operations  
   - `push(node)`: if node→rev is true, swap left/right, push rev flag to children, clear node→rev.  
   - `pull(node)`: recalculate node→size = 1 + size(left) + size(right).  
   - `split(t, k)`: split tree t into (L, R) so that L contains the first k elements in in-order, R the rest. Implementation: push(t), let Lsize = size(t.left). If k ≤ Lsize, recurse left; else recurse right with k−Lsize−1.  
   - `merge(A, B)`: push both, compare priorities: the root is whichever has higher priority; merge its right (or left) child appropriately; then pull.  

4. Reversing a range [l, r]:  
   - Split root into (A, BC) at l−1.  
   - Split BC into (B, C) at (r−l+1).  
   - Toggle B→rev flag.  
   - Merge back: root = merge(A, merge(B, C)).  

5. Final output  
   - In-order traversal of the treap.  

Time per split/merge is O(log N), so total O((N + M) log N) with very high probability.

3. Provided C++ solution with detailed comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

// Empty monoid type since we only need to store keys, not any extra data.
struct EmptyMonoid {
    static EmptyMonoid merge(EmptyMonoid a, EmptyMonoid b) {
        return EmptyMonoid();
    }
};

// Lazy tag for reversing a subtree.
template<class T>
struct ReverseLazy {
    bool should_reverse;     // if true, this subtree needs to be reversed
    ReverseLazy() : should_reverse(false) {}

    // Apply the lazy flag to a node
    template<class KeyT, class MergeFunc, uint64_t (*rng)()>
    void apply_lazy(struct TreapNode<KeyT, T, MergeFunc, ReverseLazy, rng>* node) {
        if (!node || !should_reverse) return;
        // swap children
        swap(node->left, node->right);
        // propagate flag to children
        if (node->left)  node->left->lazy.should_reverse  ^= true;
        if (node->right) node->right->lazy.should_reverse ^= true;
        // clear this node's flag
        should_reverse = false;
    }
};

template<
    class KeyT,
    class T,
    T (*merge_func)(T, T),
    class LazyT,
    uint64_t (*rng)()
>
struct TreapNode {
    KeyT key;               // the “value” stored in this node
    T data;                 // monoid data for this node (unused)
    T subtree;              // merged data of the subtree
    uint64_t prior;         // random priority
    size_t size;            // size of subtree
    TreapNode *left, *right;
    LazyT lazy;             // lazy propagation tag

    // constructor
    TreapNode(KeyT _key, T _data)
        : key(_key), data(_data), subtree(_data),
          prior(rng()), size(1),
          left(nullptr), right(nullptr), lazy() {}

    // recalc subtree data & size from children
    void pull() {
        subtree = data;
        size = 1;
        if (left) {
            left->push();                              // ensure child's lazy applied
            subtree = merge_func(left->subtree, subtree);
            size += left->size;
        }
        if (right) {
            right->push();
            subtree = merge_func(subtree, right->subtree);
            size += right->size;
        }
    }
    // apply this node's lazy tag
    void push() {
        lazy.apply_lazy(this);
    }
    // helper
    friend void push_lazy(TreapNode* t) {
        if (t) t->push();
    }

    // split by in-order size: left part has first 'cnt' elements
    friend pair<TreapNode*, TreapNode*> split_by_size(TreapNode* t, size_t cnt) {
        if (!t) return {nullptr, nullptr};
        t->push();  // apply pending reversal if any
        size_t left_sz = t->left ? t->left->size : 0;
        if (left_sz >= cnt) {
            // we want all of the split in the left subtree
            auto [L, R] = split_by_size(t->left, cnt);
            t->left = R;
            t->pull();
            return {L, t};
        } else {
            // some goes to left, some to right
            auto [L, R] = split_by_size(t->right, cnt - left_sz - 1);
            t->right = L;
            t->pull();
            return {t, R};
        }
    }

    // merge two treaps, all keys in l precede those in r
    friend TreapNode* merge(TreapNode* l, TreapNode* r) {
        push_lazy(l);
        push_lazy(r);
        if (!l || !r) return l ? l : r;
        if (l->prior > r->prior) {
            l->right = merge(l->right, r);
            l->pull();
            return l;
        } else {
            r->left = merge(l, r->left);
            r->pull();
            return r;
        }
    }
};

// wrapper treap type with our merge & lazy
template<class KeyT, class T, T (*merge_func)(T, T), class LazyT>
class Treap {
  public:
    using Node = TreapNode<KeyT, T, merge_func, LazyT, Treap::rng>;

    Node* root;

    Treap(): root(nullptr) {}

    Treap(const vector<pair<KeyT,T>>& arr) {
        build_cartesian_tree(arr);
    }

    // builds treap in O(N) from sorted-by-key array using a stack
    void build_cartesian_tree(const vector<pair<KeyT,T>>& a) {
        vector<Node*> st;
        // helper to pop stack and link
        auto recycle = [&](Node* tail) {
            Node* top = st.back(); st.pop_back();
            top->right = tail;
            return top;
        };
        for (auto &p : a) {
            Node* cur = new Node(p.first, p.second);
            Node* last = nullptr;
            while (!st.empty() && st.back()->prior < cur->prior) {
                last = recycle(last);
            }
            cur->left = last;
            st.push_back(cur);
        }
        root = nullptr;
        while (!st.empty()) {
            root = recycle(root);
        }
        pull_all(root);
    }

    // ensure all sizes/subtrees are correct
    void pull_all(Node* t) {
        if (!t) return;
        t->push();
        pull_all(t->left);
        pull_all(t->right);
        t->pull();
    }

    // single insert by key and data
    void insert(KeyT key, T data) {
        Node* nd = new Node(key, data);
        insert_in(root, nd);
    }

    // merges two treaps by in-order
    friend Treap merge_treaps(Treap l, Treap r) {
        Treap res;
        res.root = merge(l.root, r.root);
        return res;
    }

  private:
    // standard randomized insert
    static void insert_in(Node*& t, Node* it) {
        if (!t) {
            t = it;
        } else {
            t->push();
            if (it->prior > t->prior) {
                auto [L, R] = split_by_size(t, /*rank of it->key*/ it->key - 1);
                it->left = L;
                it->right = R;
                t = it;
            } else {
                insert_in(it->key < t->key ? t->left : t->right, it);
            }
        }
        if (t) t->pull();
    }
};

using MyTreap = Treap<int, EmptyMonoid, EmptyMonoid::merge, ReverseLazy<EmptyMonoid>>;

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

    int N, M;
    cin >> N >> M;

    // build initial array [(1,empty), (2,empty), ... , (N,empty)]
    vector<pair<int,EmptyMonoid>> init;
    init.reserve(N);
    for(int i = 1; i <= N; i++){
        init.emplace_back(i, EmptyMonoid());
    }
    MyTreap treap(init);

    // process M reversals
    for(int i = 0; i < M; i++){
        int l, r;
        cin >> l >> r;
        // split at l-1: [1..l-1] and [l..N]
        auto [A, BC] = TreapNode<int,EmptyMonoid,EmptyMonoid::merge,ReverseLazy<EmptyMonoid>,MyTreap::rng>
                        ::split_by_size(treap.root, l-1);
        // split BC at (r-l+1): B=[l..r], C=[r+1..N]
        auto [B, C]  = TreapNode<int,EmptyMonoid,EmptyMonoid::merge,ReverseLazy<EmptyMonoid>,MyTreap::rng>
                        ::split_by_size(BC, r-l+1);
        // mark B for reversal
        if(B) B->lazy.should_reverse ^= true;
        // reassemble
        treap.root = merge(A, merge(B, C));
    }

    // output by in-order traversal
    vector<int> ans;
    ans.reserve(N);
    function<void(MyTreap::Node*)> dfs = [&](MyTreap::Node* t){
        if (!t) return;
        t->push();
        dfs(t->left);
        ans.push_back(t->key);
        dfs(t->right);
    };
    dfs(treap.root);

    for (int x : ans) cout << x << ' ';
    cout << '\n';
    return 0;
}
```

4. Python solution with detailed comments  
```python
import sys
import threading
import random
sys.setrecursionlimit(10**7)

def main():
    input_data = sys.stdin.read().split()
    it = iter(input_data)
    n = int(next(it))
    m = int(next(it))

    # Node of an implicit treap
    class Node:
        __slots__ = ('val','prio','left','right','size','rev')
        def __init__(self, v):
            self.val = v
            self.prio = random.randrange(1<<30)
            self.left = None
            self.right = None
            self.size = 1
            self.rev = False

    # recalc size
    def pull(t):
        if not t: return
        t.size = 1
        if t.left:  t.size += t.left.size
        if t.right: t.size += t.right.size

    # push down lazy reverse flag
    def push(t):
        if t and t.rev:
            t.rev = False
            # swap children
            t.left, t.right = t.right, t.left
            # toggle children flags
            if t.left:  t.left.rev  ^= True
            if t.right: t.right.rev ^= True

    # split treap t into (L, R),
    # where L has first k elements
    def split(t, k):
        if not t:
            return (None, None)
        push(t)
        left_size = t.left.size if t.left else 0
        if k <= left_size:
            # all desired in left subtree
            L, R = split(t.left, k)
            t.left = R
            pull(t)
            return (L, t)
        else:
            # some in left, current, some in right
            L, R = split(t.right, k - left_size - 1)
            t.right = L
            pull(t)
            return (t, R)

    # merge two treaps L and R
    def merge(L, R):
        if not L or not R:
            return L or R
        # push flags before comparing
        push(L); push(R)
        if L.prio > R.prio:
            L.right = merge(L.right, R)
            pull(L)
            return L
        else:
            R.left = merge(L, R.left)
            pull(R)
            return R

    # build initial treap by successive merges
    root = None
    for x in range(1, n+1):
        node = Node(x)
        root = merge(root, node)

    # process reversals
    for _ in range(m):
        l = int(next(it))
        r = int(next(it))
        # split into A=[1..l-1], BC=[l..n]
        A, BC = split(root, l-1)
        # split BC into B=[l..r], C=[r+1..n]
        B, C  = split(BC, r-l+1)
        # toggle reverse flag on B
        if B:
            B.rev ^= True
        # reassemble
        root = merge(merge(A, B), C)

    # in-order traversal to collect result
    res = []
    def inorder(t):
        if not t: return
        push(t)
        inorder(t.left)
        res.append(t.val)
        inorder(t.right)
    inorder(root)
    print(' '.join(map(str, res)))

if __name__ == "__main__":
    threading.Thread(target=main).start()
```

5. Compressed editorial  
Use an implicit treap with random priorities, storing subtree sizes and a lazy “reverse” bit. To reverse [l,r], split at l−1 and at r, toggle the lazy bit on the middle treap, then merge back. Each operation costs O(log N), total O((N+M) log N).