BaoBao is traveling along a line with infinite length.

At the beginning of his trip, he is standing at position 0. At the beginning of each second, if he is standing at position \(x\), with \(\frac{1}{4}\) probability he will move to position \((x-1)\), with \(\frac{1}{4}\) probability he will move to position \((x+1)\), and with \(\frac{1}{2}\) probability he will stay at position \(x\). Positions can be positive, 0, or negative.

DreamGrid, BaoBao's best friend, is waiting for him at position \(m\). BaoBao would like to meet DreamGrid at position \(m\) after exactly \(n\) seconds. Please help BaoBao calculate the probability he can get to position \(m\) after exactly \(n\) seconds.

It's easy to show that the answer can be represented as \(\frac{P}{Q}\), where \(P\) and \(Q\) are coprime integers, and \(Q\) is not divisible by \(10^9+7\). Please print the value of \(PQ^{-1}\) modulo \(10^9+7\), where \(Q^{-1}\) is the multiplicative inverse of \(Q\) modulo \(10^9+7\).

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