Professor GukiZ was playing with arrays again and accidentally discovered new function, which he called GukiZiana. For given array a, indexed with integers from 1 to n, and number y, GukiZiana(a, y) represents maximum value of j - i, such that aj = ai = y. If there is no y as an element in a, then GukiZiana(a, y) is equal to - 1. GukiZ also prepared a problem for you. This time, you have two types of queries:
- First type has form 1 l r x and asks you to increase values of all ai such that l ≤ i ≤ r by the non-negative integer x.
- Second type has form 2 y and asks you to find value of GukiZiana(a, y).
For each query of type 2, print the answer and make GukiZ happy!
The first line contains two integers n, q (1 ≤ n ≤ 5 * 105, 1 ≤ q ≤ 5 * 104), size of array a, and the number of queries.
The second line contains n integers a1, a2, ... an (1 ≤ ai ≤ 109), forming an array a.
Each of next q lines contain either four or two numbers, as described in statement:
If line starts with 1, then the query looks like 1 l r x (1 ≤ l ≤ r ≤ n, 0 ≤ x ≤ 109), first type query.
If line starts with 2, then th query looks like 2 y (1 ≤ y ≤ 109), second type query.
For each query of type 2, print the value of GukiZiana(a, y), for y value for that query.
4 3 1 2 3 4 1 1 2 1 1 1 1 1 2 3
2
2 3 1 2 1 2 2 1 2 3 2 4
0 -1 【分析】给出一个数组a,然后q次操作,1 l r x表示将区间[l,r]所有数+1 ; 2 y表示询问这个数组中最左边的y和最右边的y的下标差为多少。 可分块做。将所有块里面的数排序,(便于查找要查询的y)。对于更新操作,单独处理左端点和右端点的块,改变a[i];然后对于中间跨过的块,用lazy数组标记add值。 对于查询操作,依次遍历每个块,二分查找y的位置即可。
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