For 1 \leq i \leq 215 let a_{i}=\frac{1}{2^{i}} and a_{216}=\frac{1}{2^{215}}. Let x_{1}, x_{2}, \ldots, x_{216} be positive real numbers such that
\sum_{i=1}^{216} x_{i}=1 \text { and } \sum_{1 \leq i<j \leq 216} x_{i} x_{j}=\frac{107}{215}+\sum_{i=1}^{216} \frac{a_{i} x_{i}^{2}}{2\left(1-a_{i}\right)}
The maximum possible value of x_{2}=\frac{m}{n}, where m and n are relatively prime positive integers. Find m+n.