codeforces 1006 F(折半搜索)

F. Xor-Paths

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There is a rectangular grid of size n,m) meeting the following constraints:

• You can move to the right or to the bottom only. Formally, from the cell (i+1,j). The target cell can't be outside of the grid.
• The xor of all the numbers on the path from the cell (

Find the number of such paths in the given grid.

Input

The first line of the input contains three integers k.

The next 0≤ai,j≤1018).

Output

Print one integer — the number of paths from (k.

Examples

input

Copy

3 3 11
2 1 5
7 10 0
12 6 4

output

Copy

3

input

Copy

3 4 2
1 3 3 3
0 3 3 2
3 0 1 1

output

Copy

5

input

Copy

3 4 1000000000000000000
1 3 3 3
0 3 3 2
3 0 1 1

output

Copy

0

Note

All the paths from the first example:

• (1,1)→(2,1)→(3,1)→(3,2)→(3,3);
• (1,1)→(2,1)→(2,2)→(2,3)→(3,3);
• (1,1)→(1,2)→(2,2)→(3,2)→(3,3).

All the paths from the second example:

• (1,1)→(2,1)→(3,1)→(3,2)→(3,3)→(3,4);
• (1,1)→(2,1)→(2,2)→(3,2)→(3,3)→(3,4);
• (1,1)→(2,1)→(2,2)→(2,3)→(2,4)→(3,4);
• (1,1)→(1,2)→(2,2)→(2,3)→(3,3)→(3,4);
• (1,1)→(1,2)→(1,3)→(2,3)→(3,3)→(3,4)

 

/*
暴搜2^(n+m)
折半搜索 
*/
#include<bits/stdc++.h>

#define N 27
#define ll long long

using namespace std;
ll n,m,k,ans,flag;
ll a[N][N];
map<ll,ll>M[N][N];

inline ll read()
{
     ll x=0,f=1;char c=getchar();
     while(c>'9'||c<'0'){if(x=='-')f=-1;c=getchar();}
     while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
     return x*f;
}

void dfs(int dep,int x,int y,ll sta)
{
    if(x<1 || x>n || y<1 || y>m) return;
    if(!flag) sta^=a[x][y];
    if(x+y==dep)
    {
        if(!flag){M[x][y][sta]++;return;}
        else{ans+=M[x][y][k^sta];return;}
    }
    if(!flag){
        dfs(dep,x+1,y,sta);dfs(dep,x,y+1,sta);
    }
    else{
        sta^=a[x][y];
        dfs(dep,x-1,y,sta);dfs(dep,x,y-1,sta);
    }
}

int main()
{
    n=read();m=read();k=read();
    for(int i=1;i<=n;i++) for(int j=1;j<=m;j++)
    a[i][j]=read();
    flag=0;dfs((n+m+2)/2,1,1,0);
    flag=1;dfs((n+m+2)/2,n,m,0);
    printf("%lld\n",ans);
    return 0;
}