Inspired by the tv series Stranger Things, bear Limak is going for a walk between two mirror worlds.

There are two perfect binary trees of height H, each with the standard numeration of vertices from 1 to 2^H-1. The root is 1 and the children of x are 2 \cdot x and 2 \cdot x + 1.

Let L denote the number of leaves in a single tree, L = 2^{H-1}.

You are given a permutation P_1, P_2, \ldots, P_L of numbers 1 through L. It describes L special edges that connect leaves of the two trees. There is a special edge between vertex L+i-1 in the first tree and vertex L+P_i-1 in the second tree.

graph for sample1

drawing for the first sample test, permutation P = (2, 3, 1, 4), special edges in green

Let's define product of a cycle as the product of numbers in its vertices. Compute the sum of products of all simple cycles that have exactly two special edges, modulo (10^9+7).

A simple cycle is a cycle of length at least 3, without repeated vertices or edges.

Constraints

* 2 \leq H \leq 18
* 1 \leq P_i \leq L where L = 2^{H-1}
* P_i \neq P_j (so this is a permutation)

Input

Input is given from Standard Input in the following format (where L = 2^{H-1}).


H
P_1 P_2 \cdots P_L


Output

Compute the sum of products of simple cycles that have exactly two special edges. Print the answer modulo (10^9+7).

Examples

Input

3
2 3 1 4


Output

121788


Input

2
1 2


Output

36


Input

5
6 14 15 7 12 16 5 4 11 9 3 10 8 2 13 1


Output

10199246
#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N=300005,mod=1000000007;
int n,h,a[N],x[N],y[N],s[N],d[N],ans;
vector<int>v;
int qp(int a,int b)
{
	int ans=1;
	while(b)
	{
		if(b&1)
			ans=ans*a%mod;
		a=a*a%mod;
		b>>=1;
	}
	return ans;
}
void pre(int k,int l,int r)
{
	if(l==r)
	{
		x[k]=y[k]=l;
		return;
	}
	s[k<<1]=s[k]*(k<<1)%mod;
	s[k<<1|1]=s[k]*(k<<1|1)%mod;
	int mid=l+r>>1;
	pre(k<<1,l,mid);
	pre(k<<1|1,mid+1,r);
	x[k]=x[k<<1];
	y[k]=y[k<<1|1];
}
void clear(int k)
{
	d[k]=0;
	if(k==1)
		return;
	clear(k/2);
}
void rec1(int k,int t)
{
	t=t*k%mod;
	d[k]=(d[k]+t)%mod;
	if(k==1)
		return;
	rec1(k/2,t);
}
void rec2(int k,int t,int l)
{
	t=t*k%mod;
	ans=(ans+t*(d[k^1])%mod*(k/2)%mod*qp(l,mod-2)%mod)%mod;
	if(k==1)
		return;
	rec2(k/2,t,l);
}
signed main()
{
	ios::sync_with_stdio(0);
	cin>>h;
	n=(1<<h)-1;
	for(int i=1;i<=n/2+1;i++)
		cin>>a[i];
	s[1]=1;
	pre(1,1,n/2+1);
	for(int i=1;i<=n/2;i++)
	{	
		v.clear();
		for(int j=x[i<<1];j<=y[i<<1];j++)
		{
			rec1(a[j]+n/2,s[j+n/2]);
			v.push_back(a[j]+n/2);
		}
		for(int j=x[i<<1|1];j<=y[i<<1|1];j++)
			rec2(a[j]+n/2,s[j+n/2],i==1?1:s[i]*s[i/2]%mod);
		for(auto j:v)
			clear(j);
	}	
	cout<<ans<<endl;
	return 0;
}
