B'There are n pearls in a row. Let 's enumerate them with integers from 1 to n from the left to the right. The pearl number i has the type ai. Let 's call a sequence of consecutive pearls a segment. Let 's call a segment good if it contains two pearls of the same type. Split the row of the pearls to the maximal number of good segments. Note that each pearl should appear in exactly one segment of the partition. As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java. The first line contains integer n (1 xe2 x80 x89 xe2 x89 xa4 xe2 x80 x89n xe2 x80 x89 xe2 x89 xa4 xe2 x80 x893 xc2 xb7105) -- the number of pearls in a row. The second line contains n integers ai (1 xe2 x80 x89 xe2 x89 xa4 xe2 x80 x89ai xe2 x80 x89 xe2 x89 xa4 xe2 x80 x89109) xe2 x80 x93 the type of the i-th pearl. On the first line print integer k -- the maximal number of segments in a partition of the row. Each of the next k lines should contain two integers lj, xe2 x80 x89rj (1 xe2 x80 x89 xe2 x89 xa4 xe2 x80 x89lj xe2 x80 x89 xe2 x89 xa4 xe2 x80 x89rj xe2 x80 x89 xe2 x89 xa4 xe2 x80 x89n) -- the number of the leftmost and the rightmost pearls in the j-th segment. Note you should print the correct partition of the row of the pearls, so each pearl should be in exactly one segment and all segments should contain two pearls of the same type. If there are several optimal solutions print any of them. You can print the segments in any order. If there are no correct partitions of the row print the number "-1".'...