Chubby Yang is studying linear equations right now. He came up with a nice problem. In the problem you are given an *n* × *n* matrix *W*, consisting of integers, and you should find two *n* × *n* matrices *A* and *B*, all the following conditions must hold: 
 -  *A*<sub class="lower-index">*ij*</sub> = *A*<sub class="lower-index">*ji*</sub>, for all *i*, *j* (1 ≤ *i*, *j* ≤ *n*); -  *B*<sub class="lower-index">*ij*</sub> =  - *B*<sub class="lower-index">*ji*</sub>, for all *i*, *j* (1 ≤ *i*, *j* ≤ *n*); -  *W*<sub class="lower-index">*ij*</sub> = *A*<sub class="lower-index">*ij*</sub> + *B*<sub class="lower-index">*ij*</sub>, for all *i*, *j* (1 ≤ *i*, *j* ≤ *n*). 
Can you solve the problem?
The first line contains an integer *n* (1 ≤ *n* ≤ 170). Each of the following *n* lines contains *n* integers. The *j*-th integer in the *i*-th line is *W*<sub class="lower-index">*ij*</sub> (0 ≤ |*W*<sub class="lower-index">*ij*</sub>| &lt; 1717).
#include <iostream>
using namespace std;
int main(){
	int n;
	cin>>n;
	int data[n][n];
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			cin>>data[i][j];
		}
	}
	float a[n][n], b[n][n];
	for(int i=0;i<n;i++){
		a[i][i] = data[i][i];
		b[i][i] = 0;
	}
	for(int i=0;i<n;i++){
		for(int j=i+1;j<n;j++){
			a[i][j] = a[j][i] = .5*(data[i][j]+data[j][i]);
			b[i][j] = data[i][j]-a[i][j];
			b[j][i] = -b[i][j];
		}
	}
	cout.precision(2);
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			cout<<fixed<<float(a[i][j])<<" ";
		}
		cout<<endl;
	}
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			cout<<fixed<<b[i][j]<<" ";
		}
		cout<<endl;
	}
	return 0;
}
