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 mod 1000000007
using namespace std;
typedef long long int ll;
int n,a[(1<<19)+1];
bool leaf[(1<<19)+1];
ll ans,b[(1<<19)+1],c[(1<<19)+1];
vector<int>wait,tmp;
void dfs(int x,ll s,int g)
{
	tmp.push_back(x);
	if(x>=n)
	{
		wait.push_back(a[x-n+1]+n-1);
		if(g)
			b[a[x-n+1]+n-1]=s*(a[x-n+1]+n-1)%mod;
		else
			c[a[x-n+1]+n-1]=s*(a[x-n+1]+n-1)%mod;
		tmp.push_back(a[x-n+1]+n-1);
		return;
	}
	assert(s>=0);
	dfs(x<<1,s*(x<<1)%mod,g);
	dfs(x<<1|1,s*(x<<1|1)%mod,g);
}
inline void solve(ll x)
{
	dfs(x<<1,x<<1,0);
	for(int i=0;i<wait.size();++i)
	{
		int u=wait[i];
		ll now=c[u];
		tmp.push_back(u);
		u>>=1;
		while(u)
		{
			tmp.push_back(u);
			now=now*u%mod;
			c[u]=(c[u]+now)%mod;
			u>>=1;
		}
	}
	wait.clear();
	dfs(x<<1|1,(x<<1|1)*x%mod,1);
	for(int i=0;i<wait.size();++i)
	{
		int u=wait[i];
		ll now=b[u];
		while(u!=1)
		{
			ans=(ans+now*c[u^1]%mod*(u>>1)%mod)%mod;
			u>>=1;
			now=now*u%mod;
		}
	}
	wait.clear();
	for(int i=0;i<tmp.size();++i)
		c[tmp[i]]=b[tmp[i]]=0;
	tmp.clear();
}
int main()
{
	ios::sync_with_stdio(false);
	cin>>n;
	n=1<<(n-1);
	for(int i=1;i<=n;++i)
		cin>>a[i];
	for(int i=1;i<=n-1;++i)
		solve(i);
	cout<<(ans%mod+mod)%mod<<endl;
	return 0;
}
