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>
using namespace std;
const int N=3e5+10;
const int mod=1e9+7;
typedef vector<int> vi;
#define pb push_back

int prd[N],iprd[N],inv[N],n,col[N],m,p[N],val[N],sum,ans,f[N][2];
bool vis[N];

void access(int x,bool op) {
	while(x) vis[x]=op,x>>=1;
}

void dfs(int u,int c) {
	if(u>m) {
		col[p[u]]=c;
		val[p[u]]=1ll*prd[p[u]]*prd[u]%mod;
		access(p[u],1);
		return;
	}
	dfs(u<<1,c);
	dfs(u<<1|1,c);
}

void clear(int u) {
	if(u>m) {
		access(p[u],0);
		return;
	}
	clear(u<<1);
	clear(u<<1|1);
}

void dfs2(int u) {
	f[u][0]=f[u][1]=0;
	if(!vis[u]) return;
	if(u>m) {
		f[u][col[u]]=val[u];
		return;
	}
	dfs2(u<<1);
	dfs2(u<<1|1);
	int tmp=(1ll*f[u<<1][0]*f[u<<1|1][1]+1ll*f[u<<1][1]*f[u<<1|1][0])%mod;
	sum=(sum+1ll*tmp*iprd[u]%mod*iprd[u>>1])%mod;
	f[u][0]=(f[u<<1][0]+f[u<<1|1][0])%mod;
	f[u][1]=(f[u<<1][1]+f[u<<1|1][1])%mod;
}

int main() {
	ios::sync_with_stdio(0);cin.tie(0);
	cin>>n;n=(1<<n)-1;m=n/2;
	for(int i=m+1;i<=n;i++) cin>>p[i],p[i]+=m;
	inv[1]=1;for(int i=2;i<=n;i++) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
	prd[0]=iprd[0]=1;
	for(int i=1;i<=n;i++) {
		prd[i]=1ll*prd[i>>1]*i%mod;
		iprd[i]=1ll*iprd[i>>1]*inv[i]%mod;
	}
	for(int i=1;i<=m;i++) {
		dfs(i<<1,0);
		dfs(i<<1|1,1);
		sum=0;dfs2(1);
		ans=(ans+1ll*sum*iprd[i]%mod*iprd[i>>1])%mod;
		clear(i);
	}
	cout<<ans;
	return 0;
}
