Educational Codeforces Round 32 F. Connecting Vertices

F. Connecting Vertices

time limit per test4 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

There are n points marked on the plane. The points are situated in such a way that they form a regular polygon (marked points are its vertices, and they are numbered in counter-clockwise order). You can draw n - 1 segments, each connecting any two marked points, in such a way that all points have to be connected with each other (directly or indirectly).

But there are some restrictions. Firstly, some pairs of points cannot be connected directly and have to be connected undirectly. Secondly, the segments you draw must not intersect in any point apart from the marked points (that is, if any two segments intersect and their intersection is not a marked point, then the picture you have drawn is invalid).

How many ways are there to connect all vertices with n - 1 segments? Two ways are considered different iff there exist some pair of points such that a segment is drawn between them in the first way of connection, but it is not drawn between these points in the second one. Since the answer might be large, output it modulo 109 + 7.

Input

The first line contains one number n (3 ≤ n ≤ 500) — the number of marked points.

Then n lines follow, each containing n elements. ai, j (j-th element of line i) is equal to 1 iff you can connect points i and j directly (otherwise ai, j = 0). It is guaranteed that for any pair of points ai, j = aj, i, and for any point ai, i = 0.

Output

Print the number of ways to connect points modulo 109 + 7.

Examples

input

3

0 0 1

0 0 1

1 1 0

output

1

input

4

0 1 1 1

1 0 1 1

1 1 0 1

1 1 1 0

output

12

input

3

0 0 0

0 0 1

0 1 0

output

0

题意就是给出一个n正边形，给你一个n*n的矩阵，a[i][j]为1代表i与j可以直接连接，不然代表i，j不能直接连接，问有多少种可能用n-1条不相交的边使得这n个点联通。
开d

和f

，d[i][j]代表i与j相连之后i到j之间的点有多少种连接方式，f[i][j]代表i到h任意相连得到的连接方式。
d[i][j]就相当于把i到j分成两个部分，再把i和j相连，d[i][j]就是（i<=k<j）f[i][k]*f[k+1][j]
对于f状态，对于i，j之间，枚举i到j属于一个集合，即i与k相连，再连接其他边。f[i][j]就是（i<k<=j）d[i][k]*f[k][j].

#include<bits/stdc++.h>
#define mod 1000000007
#define Maxn 507
using namespace std;

int n,a[Maxn][Maxn],d[Maxn][Maxn],f[Maxn][Maxn];

int main()
{
scanf("%d",&n);
for(int i=0;i<n;i++)
for (int j=0;j<n;j++)
scanf("%d",&a[i][j]);

for(int i=0;i<n;i++)
d[i][i]=1,f[i][i]=1;

for(int len=2;len<=n;len++)
{
for(int i=0;i<n;i++)
{
int j=(i+len-1)%n;
for(int k=i;k!=j;k=(k+1)%n)
{
if(a[i][j])
d[i][j]=(d[i][j]+1LL*f[i][k]*f[(k+1)%n][j])%mod;
f[i][j]=(f[i][j]+1LL*d[i][(k+1)%n]*f[(k+1)%n][j])%mod;
}
//cout << i << ' '<< j << ' '<< d[i][j] << ' '<< f[i][j] << endl;
}
}
printf("%d\n",f[0][n-1]);
return 0;
}