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 <bits/stdc++.h>
#define ll long long
#define ld long double
using namespace std;

void solve(){
    int n;cin>>n;
    vector<vector<ld>> w(n,vector<ld>(n));
    vector<vector<ld>> w_t(n,vector<ld>(n));
    for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
            cin>>w[i][j];
            w_t[j][i]=w[i][j];
            }
    }
    // first is the symmetric matrix part 0.5(w+w_t)
    cout<<setprecision(6);
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            cout<<0.5*(w[i][j]+w_t[i][j])<<' ';
        }
        cout<<'\n';
    }
    // second is antisymmetric matrix
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            cout<<0.5*(w[i][j]-w_t[i][j])<<' ';
        }
        cout<<'\n';
    }
}




int main(void){
    ios::sync_with_stdio(0);
    cin.tie(0);
    solve();
    return 0;
}
