Codeforces Round #198 (Div. 1) D. Iahub and Xors 二维树状数组*

D. Iahub and Xors

 

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:

1. Query(x0, y0, x1, y1): print the xor sum of the elements of the submatrix (x0, y0, x1, y1).
2. Update(x0, y0, x1, y1, v): each element from submatrix (x0, y0, x1, y1) gets xor-ed by value v.

Input

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 thei-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.

Output

For each query operation, output on a new line the result.

Examples

input

3 5
2 1 1 2 2 1
2 1 3 2 3 2
2 3 1 3 3 3
1 2 2 3 3
1 2 2 3 2

output

3
2

Note

After the first 3 operations, the matrix will look like this:

1 1 2
1 1 2
3 3 3

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.

题意：

　　给你一个n×n的矩阵；初始时所有位置的值都为0；

　　m次操作：两种操作，一种是将某个矩形区域的值异或v，另一种是求一个矩形区域的异或和。

题解：

　　想办法维护(1,1)(x,y)的异或和，那么矩形区域的异或和就可以根据四个位置的异或和求出。由于异或操作比较特殊，在一个区域进行异或操作，那么这个区域的某个点的(1,1)(x,y)的异或和要么要异或上v，要么异或上0。这样，我们就可以利用二维树状数组，只进行单点更新就完成更新矩形区域的操作，另外，由于区域是否被更新跟奇偶性有关，因此跟据奇偶性来维护4种情况的树状数组。

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double Pi = acos(-1.0);
const int N = 1e3+, M = 1e5+, mod = 1e9+, inf = 2e9;

int n,m,mp[N][N];
LL C[][N][N];

int get(int x,int y) {
        int res = ;
        if(x&) res += ;
        if(y&) res += ;
        return res;
}
LL ask(int x,int y) {
        LL s = ;
        int wh = get(x,y);
         for(int i = x; i; i-= i & (-i))
            for(int j = y; j; j -= j & (-j)) s^=C[wh][i][j];
        return s;
}
void update(int x,int y,LL v) {
        int wh = get(x,y);
        for(int i = x; i < N; i+= i&(-i))
            for(int j = y; j < N; j += j&(-j)) C[wh][i][j] ^= v;
}
int main() {
    scanf("%d%d",&n,&m);
    for(int i = ; i <= m; ++i) {
        int op,x1,x2,y1,y2;
        LL v;
        scanf("%d%d%d%d%d",&op,&x1,&y1,&x2,&y2);
        if(op == ) {
           LL a = ask(x2,y2);
           LL b = , c = , d = ;
           if(y1 > ) b = ask(x2,y1-);
           if(x1 > ) c = ask(x1-,y2);
           if(x1 >  && y1 > ) d = ask(x1-,y1-);
           printf("%I64d\n",a^b^d^c);
        } else {
            scanf("%I64d",&v);
            update(x2+,y2+,v);
            update(x2+,y1,v);
            update(x1,y2+,v);
            update(x1,y1,v);
        }
    }
    return ;
}