D. Iahub and Xors
data structures
Iahub does not like background stories, so he'll tell you exactly what this problem asks you for.You are given a matrix a with n rows and n columns. Initially, all values of the matrix are zeros. Both rows and columns are 1-based, that is rows are numbered 1, 2, ..., n and columns are numbered 1, 2, ..., n. Let's denote an element on the i-th row and j-th column as ai, j.We will call a submatrix (x0, y0, x1, y1) such elements ai, j for which two inequalities hold: x0 ≤ i ≤ x1, y0 ≤ j ≤ y1.Write a program to perform two following operations: Query(x0, y0, x1, y1): print the xor sum of the elements of the submatrix (x0, y0, x1, y1). Update(x0, y0, x1, y1, v): each element from submatrix (x0, y0, x1, y1) gets xor-ed by value v.
The first line contains two integers: n (1 ≤ n ≤ 1000) and m (1 ≤ m ≤ 105). The number m represents the number of operations you need to perform. Each of the next m lines contains five or six integers, depending on operation type. If the i-th operation from the input is a query, the first number from i-th line will be 1. It will be followed by four integers x0, y0, x1, y1. If the i-th operation is an update, the first number from the i-th line will be 2. It will be followed by five integers x0, y0, x1, y1, v. It is guaranteed that for each update operation, the following inequality holds: 0 ≤ v < 262. It is guaranteed that for each operation, the following inequalities hold: 1 ≤ x0 ≤ x1 ≤ n, 1 ≤ y0 ≤ y1 ≤ n.
For each query operation, output on a new line the result.
After the first 3 operations, the matrix will look like this: 1 1 21 1 23 3 3The fourth operation asks us to compute 1 xor 2 xor 3 xor 3 = 3.The fifth operation asks us to compute 1 xor 3 = 2.