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 a i , j . We will call a submatrix ( x 0 , y 0 , x 1 , y 1 ) such elements a i , j for which two inequalities hold: x 0 ≤ i ≤ x 1 , y 0 ≤ j ≤ y 1 . Write a program to perform two following operations: Query( x 0 , y 0 , x 1 , y 1 ): print the xor sum of the elements of the submatrix ( x 0 , y 0 , x 1 , y 1 ) . Update( x 0 , y 0 , x 1 , y 1 , v ): each element from submatrix ( x 0 , y 0 , x 1 , y 1 ) gets xor-ed by value v . The first line contains two integers: n ( 1 ≤ n ≤ 1000 ) and m ( 1 ≤ m ≤ 10 5 ). 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 x 0 , y 0 , x 1 , y 1 . 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 x 0 , y 0 , x 1 , y 1 , v . It is guaranteed that for each update operation, the following inequality holds: 0 ≤ v < 2 62 . It is guaranteed that for each operation, the following inequalities hold: 1 ≤ x 0 ≤ x 1 ≤ n , 1 ≤ y 0 ≤ y 1 ≤ n . For each query operation, output on a new line the result. After the first 3 operations, the matrix will look like this: The 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.