<p>Given two strings <code>a</code> and <code>b</code>, return <em>the minimum number of times you should repeat string </em><code>a</code><em> so that string</em> <code>b</code> <em>is a substring of it</em>. If it is impossible for <code>b</code>​​​​​​ to be a substring of <code>a</code> after repeating it, return <code>-1</code>.</p>

<p><strong>Notice:</strong> string <code>&quot;abc&quot;</code> repeated 0 times is <code>&quot;&quot;</code>, repeated 1 time is <code>&quot;abc&quot;</code> and repeated 2 times is <code>&quot;abcabc&quot;</code>.</p>

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

<pre>
<strong>Input:</strong> a = &quot;abcd&quot;, b = &quot;cdabcdab&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> We return 3 because by repeating a three times &quot;ab<strong>cdabcdab</strong>cd&quot;, b is a substring of it.
</pre>

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

<pre>
<strong>Input:</strong> a = &quot;a&quot;, b = &quot;aa&quot;
<strong>Output:</strong> 2
</pre>

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

<ul>
	<li><code>1 &lt;= a.length, b.length &lt;= 10<sup>4</sup></code></li>
	<li><code>a</code> and <code>b</code> consist of lowercase English letters.</li>
</ul>

class Solution {
public:
    const long long M = 1e9 + 7;
    const long long p = 31;
    int repeatedStringMatch(string a, string b) {
        vector<long long>p_pow(1e6 + 1);
        p_pow[0] = 1;
        for(int i = 1; i < 1e6; i++){
            p_pow[i] = (p_pow[i - 1] * p) % M;
        }
        long long h_b = 0;
        for(int i = 0; i < b.size(); i++){
            h_b = (h_b +(b[i] - 'a' + 1) * p_pow[i]) % M;
        }
        vector<long long>h_a({0});
        for(int i = 0; i < a.size(); i++){
            long long tmp = h_a.back();
            tmp = (tmp + (a[i] - 'a' + 1) * p_pow[i]) % M;
            h_a.push_back(tmp);
        }
        bool p = true;
        int ans = 1;
        int id = 0;
        string orig = a;

        while(p){
            if(a.size() >= 1e5 || ans > 1e4)break;
            int sz_a = a.size();
            int sz_b = b.size();
            long long hash = 0;
            while(sz_a - id >= sz_b){
                hash = (h_a[id + sz_b] - h_a[id] + M) % M;
                // cout<<hash<<" "<<(h_b * p_pow[id]) % M<<"\n";
                if(hash == (h_b * p_pow[id]) % M)return ans;
                id++;
            }

            for(int i = 0; i < orig.size(); i++){
                long long tmp = h_a.back();
                tmp = (tmp + (orig[i] - 'a' + 1) * p_pow[i + a.size()]) % M;
                h_a.push_back(tmp);
            }

            a += orig;
            // cout<<a<<" "<<id<<"\n";
            ans++;
        }
        return -1;
    }
};
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