B"East or west, home is best. That's why family reunion, the indispensable necessity of Lunar New Year celebration, is put in such a position. After the reunion dinner, Little Tommy plays a game with the family. Here is a concise introduction to this game: Obviously, every game ends after at most n xe2 x80 x89- xe2 x80 x891 descensions. Please share your solution of this game with the lowest cost. The first line contains one integer n (1 xe2 x80 x89 xe2 x89 xa4 xe2 x80 x89n xe2 x80 x89 xe2 x89 xa4 xe2 x80 x893 xc2 xb7105). The second line contains n space-separated integers p1, xe2 x80 x89p2, xe2 x80 x89..., xe2 x80 x89pn (0 xe2 x80 x89 xe2 x89 xa4 xe2 x80 x89pi xe2 x80 x89 xe2 x89 xa4 xe2 x80 x89109, xe2 x80 x89i xe2 x80 x89= xe2 x80 x891, xe2 x80 x892, xe2 x80 x89..., xe2 x80 x89n). In the first line print one integer as the number of descensions m (0 xe2 x80 x89 xe2 x89 xa4 xe2 x80 x89m xe2 x80 x89 xe2 x89 xa4 xe2 x80 x89n xe2 x80 x89- xe2 x80 x891). In the next m lines print the descensions chronologically. More precisely, in each line of the next m lines print one integer i (1 xe2 x80 x89 xe2 x89 xa4 xe2 x80 x89i xe2 x80 x89< xe2 x80 x89n) representing a descension would operate on pi and pi xe2 x80 x89+ xe2 x80 x891 such that all the descensions could be utilized from top to bottom. If there are many possible solutions to reach the minimal cost, print any of them. In the first sample, one possible best solution is , of which the cost is 1 xe2 x80 x89+ xe2 x80 x891 xe2 x80 x89= xe2 x80 x892. In the second sample, one possible best solution is , of which the cost is 1 xe2 x80 x89+ xe2 x80 x891 xe2 x80 x89+ xe2 x80 x891 xe2 x80 x89= xe2 x80 x893. "...