<p>A positive integer is <em>magical</em> if it is divisible by either <code>a</code> or <code>b</code>.</p>

<p>Given the three integers <code>n</code>, <code>a</code>, and <code>b</code>, return the <code>n<sup>th</sup></code> magical number. Since the answer may be very large, <strong>return it modulo </strong><code>10<sup>9</sup> + 7</code>.</p>

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

<pre>
<strong>Input:</strong> n = 1, a = 2, b = 3
<strong>Output:</strong> 2
</pre>

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

<pre>
<strong>Input:</strong> n = 4, a = 2, b = 3
<strong>Output:</strong> 6
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>9</sup></code></li>
	<li><code>2 &lt;= a, b &lt;= 4 * 10<sup>4</sup></code></li>
</ul>

class Solution {
public:

    int nthMagicalNumber(int n, int a, int b) {
        int A=a;
        int B=b;
        long long mod=pow(10,9)+7;
        long long left=min(a,b);
        long long  right=(long)n*min(a,b);

        while(B>0){
            int temp=A;
            A=B;
            B=temp%A;
        }
        long long lcm=(a*b)/A;
        while(left<right){
            long long m=left+(right-left)/2;
            if((m/a+m/b-m/lcm)<n) left=m+1;
            else right=m;
        }
        return left%mod;
    }
};
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