1. Abridged Problem Statement  
You have a gold chain of N links and must pawn exactly one link per day for up to N days. At the start you may break some links (each broken-open link becomes a "connector" you can use for change). On each day you hand over one or more contiguous chain segments, and the pawnbroker returns any previously pawned individual links as change, so that your net payment is exactly one link. Determine the minimum number of links you must break up front so that you can pay exactly one link per day for all N days.

2. Detailed Editorial  
Let k be the number of links you choose to break initially. Each broken link becomes a single loose ring; the remaining intact parts form k+1 connected segments of rings. You will use those k broken rings plus the k+1 segments to make your daily payments, always handing over entire segments or loose rings and getting back previously pawned rings as change so that your net is exactly one ring per day.

Our goal is to choose k as small as possible so that with k broken rings and k+1 segments you can cover all days from day 1 up to day N.

Key observation: if you break k links optimally (i.e., choose the segment lengths to maximize coverage), the maximum number of days M(k) you can pay is

  M(k) = (k+1) × 2^(k+1) − 1.

Proof sketch of the formula:  
1. For k=0 (no breaks), you have one segment of length N and no loose rings; you can only pay on day 1, so M(0)=1=1×2¹−1.  
2. Assume with k−1 breaks the coverage is M(k−1)=k·2^k−1. Adding one more break gives one extra loose ring and lets you double your "payment reach" plus add another full cycle; a short induction confirms M(k)=(k+1)·2^(k+1)−1.

We want the smallest k such that M(k) ≥ N. Once we know k, that is exactly how many links we must break.

Implementation:  
— Initialize ans=0 (current number of breaks) and x=2 (we keep x=2^(ans+1)).  
— While x*(ans+1)−1 < N, increment ans and double x.  
— Output ans.

Time complexity is O(log N) because x doubles each iteration and grows very quickly.

3. C++ Solution

```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;
};

int64_t n;

void read() { cin >> n; }

void solve() {
    // With b broken links we get b single loose links plus b + 1 intact
    // segments. The b loose links can make change for any value 1..b directly,
    // and the segments (chosen with the classic 1, 2, 4, ..., 2^b sizes) let us
    // cover everything up to (b + 1) * 2^b - 1 days, where the -1 accounts for
    // the loose links being part of the total. So we find the smallest b with
    // (b + 1) * 2^b - 1 >= n by incrementing b while the bound is still below.

    int64_t x = 2, ans = 0;
    while(x * (ans + 1) - 1 < n) {
        ans++;
        x *= 2;
    }

    cout << ans << '\n';
}

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;
}
```

4. Python Solution

```python
import sys

def main():
    data = sys.stdin.read().strip()
    if not data:
        return
    # Parse N (up to 1e16)
    n = int(data)

    # We'll track ans = number of links broken
    # and x = 2^(ans+1).  Start with ans=0 => x=2.
    ans = 0
    x = 2
    # We need the minimal ans such that (ans+1)*2^(ans+1)-1 >= n
    # Loop while we haven't reached n yet
    while x * (ans + 1) - 1 < n:
        ans += 1
        x <<= 1  # same as x = x * 2

    # Output the result
    print(ans)

if __name__ == "__main__":
    main()
```

5. Compressed Editorial  
To pay one link per day for N days with minimal breaks, note that breaking k links yields k loose rings and k+1 intact segments. With an optimal choice of segment lengths, one can cover up to M(k) = (k+1)·2^(k+1)−1 days. Thus find the smallest k for which M(k) ≥ N; that k is the answer. Iteratively maintain x = 2^(k+1) and increase k until x·(k+1)−1 ≥ N.
