<p>You are <strong>building</strong> a string <code>s</code> of length <code>n</code> <strong>one</strong> character at a time, <strong>prepending</strong> each new character to the <strong>front</strong> of the string. The strings are labeled from <code>1</code> to <code>n</code>, where the string with length <code>i</code> is labeled <code>s<sub>i</sub></code>.</p>

<ul>
	<li>For example, for <code>s = &quot;abaca&quot;</code>, <code>s<sub>1</sub> == &quot;a&quot;</code>, <code>s<sub>2</sub> == &quot;ca&quot;</code>, <code>s<sub>3</sub> == &quot;aca&quot;</code>, etc.</li>
</ul>

<p>The <strong>score</strong> of <code>s<sub>i</sub></code> is the length of the <strong>longest common prefix</strong> between <code>s<sub>i</sub></code> and <code>s<sub>n</sub></code> (Note that <code>s == s<sub>n</sub></code>).</p>

<p>Given the final string <code>s</code>, return<em> the <strong>sum</strong> of the <strong>score</strong> of every </em><code>s<sub>i</sub></code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;babab&quot;
<strong>Output:</strong> 9
<strong>Explanation:</strong>
For s<sub>1</sub> == &quot;b&quot;, the longest common prefix is &quot;b&quot; which has a score of 1.
For s<sub>2</sub> == &quot;ab&quot;, there is no common prefix so the score is 0.
For s<sub>3</sub> == &quot;bab&quot;, the longest common prefix is &quot;bab&quot; which has a score of 3.
For s<sub>4</sub> == &quot;abab&quot;, there is no common prefix so the score is 0.
For s<sub>5</sub> == &quot;babab&quot;, the longest common prefix is &quot;babab&quot; which has a score of 5.
The sum of the scores is 1 + 0 + 3 + 0 + 5 = 9, so we return 9.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;azbazbzaz&quot;
<strong>Output:</strong> 14
<strong>Explanation:</strong> 
For s<sub>2</sub> == &quot;az&quot;, the longest common prefix is &quot;az&quot; which has a score of 2.
For s<sub>6</sub> == &quot;azbzaz&quot;, the longest common prefix is &quot;azb&quot; which has a score of 3.
For s<sub>9</sub> == &quot;azbazbzaz&quot;, the longest common prefix is &quot;azbazbzaz&quot; which has a score of 9.
For all other s<sub>i</sub>, the score is 0.
The sum of the scores is 2 + 3 + 9 = 14, so we return 14.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

const int M = 1e9 + 9;
const int BASE1 = 5689;
const int BASE2 = 8861;

// Hash Function
struct Hash {
    vector<pair<long long int, long long int>> hashes, Pow;
    
    Hash(const string &s) {
        long long int n = s.size();
        hashes.assign(n + 1, {0, 0});
        Pow.assign(n + 1, {1, 1});

        for (int i = 0; i < n; i++) {
            hashes[i + 1] = {
                (hashes[i].first * BASE1 + (s[i] - 'a' + 1)) % M,
                (hashes[i].second * BASE2 + (s[i] - 'a' + 1)) % M
            };
            Pow[i + 1] = {
                (Pow[i].first * BASE1) % M,
                (Pow[i].second * BASE2) % M
            };
        }
    }

    pair<int, int> get(int l, int r) {
        l++, r++;
        long long hash1 = (hashes[r].first - (long long)(hashes[l - 1].first * Pow[r - l + 1].first) % M + M) % M;
        long long hash2 = (hashes[r].second - (long long)(hashes[l - 1].second * Pow[r - l + 1].second) % M + M) % M;
        return {hash1, hash2};
    }
};

class Solution {
public:
    long long sumScores(string s) {
        int n = s.size();
        Hash a(s);
        long long total_score = 0;

        for (int i = 1; i < n; ++i) {
            int l = 0, h = n - i, best = 0;

            while (l <= h) {
                int mid = (l + h) / 2;
                if (a.get(0, mid - 1) == a.get(i, i + mid - 1)) {
                    best = mid;
                    l = mid + 1;
                } else {
                    h = mid - 1;
                }
            }

            total_score += best;
        }

        return total_score+n;
    }
};
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