. Finally, (X) ? 1, hence d X1 (y) ? d Y (y) ? 1. Moreover d Y1 (x) ? 2|Y | + 5 ? e(X) ? e(X) + 5 ? d X (x) + 5, all, we have e(X 1 , Y 1 ) ? e(X, Y ) ? 4 ?

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