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
const int mod = 1000000007;
using namespace std;

int sum[(1<<18)+321], mulpath[(1<<18)+321], revpath[(1<<18)+321];
int p[(1<<18)+321];
int n, ans=0;
int cl[(1<<18)+321], cr[(1<<18)+321];
vector<int> cleanse;

void mul(int &x, int k)
{
	x=(x*k)%mod;
}

void add(int &x, int k)
{
	x+=k;
	if(x>mod) x-=mod;
}

int binpow(int a, int b)
{
	int ans=1, cac=a;
	while(b)
	{
		if(b&1)
		{
			ans=(ans*cac)%mod;
		}
		b>>=1;
		cac=(cac*cac)%mod;
	}
	return ans;
}

void dfs(int node)
{
	mulpath[node]=node;
	if(node>1) mulpath[node]=(mulpath[node]*mulpath[node/2])%mod;
	if(node>n/2)
	{
		cl[node]=node;
		cr[node]=node;
		return;
	}
	int ch=node*2;
	dfs(ch);
	cl[node]=cl[ch];
	ch+=1;
	dfs(ch);
	cr[node]=cr[ch];
}

void gol(int node, int org), gor(int node, int org);

void solve(int node)
{
	// cout<<node<<endl;
	for(int i=cl[node*2]; i<=cr[node*2]; i++)
	{
		gol(p[i], ((mulpath[i]*revpath[node/2])%mod*mulpath[p[i]])%mod);
	}
	for(int i=cl[node*2+1]; i<=cr[node*2+1]; i++)
	{
		gor(p[i], ((mulpath[i]*revpath[node])%mod*mulpath[p[i]])%mod);
	}
	for(auto i:cleanse)
	{
		sum[i]=0;
	}
	cleanse.clear();
}

void gol(int node, int org)
{
	while(node)
	{
		cleanse.push_back(node);
		int ins=org;
		mul(ins, revpath[node/2]);
		add(sum[node], ins);
		if(node>1)
		{
			ins=org;
			mul(ins, revpath[node/4]);
			mul(ins, node);
			add(sum[node], mod-ins);
		}
		node>>=1;
	}
}

void gor(int node, int org)
{
	while(node)
	{
		int ins=org;
		mul(ins, revpath[node]);
		mul(ins, sum[node]);
		add(ans, ins);
		node>>=1;
	}
}

signed main()
{
	cin>>n;
	n=(1<<n)-1;
	mulpath[0]=1;
	for(int i=n/2+1; i<=n; i++)
	{
		cin>>p[i];
		p[i]+=n/2;
	}
	dfs(1);
	for(int i=0; i<=n; i++)
	{
		revpath[i]=binpow(mulpath[i], mod-2);
	}
	// cout<<n<<endl;
	for(int i=1; i<=n; i++)
	{
		if(i<=n/2)
		{
			solve(i);
		}
	}
	cout<<ans;
}