#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define N 100
using namespace std;
int a[N][N];
double aa[N][N];
int n;/**********************************************************/
//求行列式的值：是所有取自不同行不同列的n个元素的乘积
int d[N];
int tmpd[N];
int dd[N];
int vis_column[N];//是否在同一列
int inverse;//逆序数的个数
int ans = 0;
void merger(int ld, int rd){if(ld >= rd) return;int mid=(ld+rd)>>1;merger(ld,mid);merger(mid+1, rd);int i=ld, j=mid+1, k=0;while(i<=mid && j<=rd){int col1 = (tmpd[i]-1)%n+1;int col2 = (tmpd[j]-1)%n+1;if(col1 < col2){dd[++k] = tmpd[i++];} else {dd[++k] = tmpd[j++];inverse += mid-i+1; }}while(i<=mid) dd[++k] = tmpd[i++];while(j<=rd) dd[++k] = tmpd[j++];memcpy(tmpd+ld, dd+1, sizeof(int)*(k));
}void determinantValue(int row){if(row>n){inverse=0; int tmp = 1;for(int k=1; k<=n; ++k){int i = (d[k]-1)/n+1;int j = (d[k]-1)%n+1;tmp *= a[i][j];}memcpy(tmpd, d, sizeof(int)*(n+1));merger(1, n);if(inverse&1) ans-=tmp;else ans+=tmp;return;}for(int col=1; col<=n; ++col)if(!vis_column[col]){vis_column[col]=1;d[row] = (row-1)*n + col;determinantValue(row+1);vis_column[col]=0;}
}
/**********************************************************//**********************************************************/
//以列主元方法，将行列式转变成上三角矩阵double determinantValueA(){int sign = 1;double ret = 0.0;for(int i=1; i<=n; ++i){double maxVal = aa[i][i];int j = i;for(int k=i+1; k<=n; ++k)//寻找这一列中的元素值的最大值 if(maxVal < aa[k][i]){maxVal = aa[k][i];j = k;}if(j!=i){sign = -sign;for(int k=1; k<=n; ++k){//交换两行 double tmp = aa[i][k];aa[i][k] = aa[j][k];                 aa[j][k] = tmp;}}if(maxVal < 1e-10)//最大值为0，也就是意味这对角线上的元素有0出现 return ret;for(int k=i+1; k<=n; ++k){double b = aa[k][i]/aa[i][i];for(int c=1; c<=n; ++c) aa[k][c] -= b*aa[i][c];}}ret = 1.0;for(int i=1; i<=n; ++i)ret*=aa[i][i];return ret;
}/**********************************************************/
int main() {
//    cin>>n;
//    for(int i=1; i<=n; ++i)
//        for(int j=1; j<=n; ++j)
//            cin>>a[i][j];
//    determinantValue(1);
//    cout<<ans<<endl;cin>>n;for(int i=1; i<=n; ++i)for(int j=1; j<=n; ++j)cin>>aa[i][j];cout<<determinantValueA()<<endl;return 0;
}
/*
3 4 5 11
5 4 9
3 2 12
-11 21 29
2 0 0 0
2 0 0
0 2 0
0 0 2
1 2 3 4
*/