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 ll long long
#define ld long double
#define db double
#define pint pair<int,int>
#define mk(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define eb(x,y) emplace_back(x,y)
#define fi first
#define se second
#define Rep(x,y,z) for(int x=y;x<=z;x++)
#define Red(x,y,z) for(int x=y;x>=z;x--)
using namespace std;
const int MAXN=(1<<18)+5,Mod=1e9+7;
char buf[1<<12],*pp1=buf,*pp2=buf,nc;int ny;
inline char gc() {return pp1==pp2&&(pp2=(pp1=buf)+fread(buf,1,1<<12,stdin),pp1==pp2)?EOF:*pp1++;}
//inline char gc(){return getchar();}
inline int read(){
	int x=0;for(ny=1;nc=gc(),(nc<48||nc>57)&&nc!=EOF;)if(nc==45)ny=-1;if(nc<0)return nc;
	for(x=nc-48;nc=gc(),47<nc&&nc<58&&nc!=EOF;x=(x<<3)+(x<<1)+(nc^48));return x*ny;
}
int n,H,p[MAXN],tag[MAXN],ans,S[MAXN],iS[MAXN];
inline void Ins(int x,int Sum){Sum=1ll*Sum*S[p[x]]%Mod;for(int t=p[x];t;t>>=1)tag[t]=(tag[t]+Sum)%Mod;}
inline void Del(int x,int Sum){Sum=1ll*Sum*S[p[x]]%Mod;for(int t=p[x];t;t>>=1)tag[t]=(tag[t]-Sum+Mod)%Mod;}
inline int Fp(int x,int k){int ans=1;for(;k;k>>=1,x=1ll*x*x%Mod)if(k&1)ans=1ll*ans*x%Mod;return ans;}
void Solve(int x,int l,int r){
	if(l==r){Ins(l,S[x]);return;}int mid=l+r>>1;
	Solve(x<<1,l,mid);
	Rep(i,mid+1,r){
		int Sum=1ll*S[i]*iS[x]%Mod*iS[x>>1]%Mod*p[i]%Mod;
		for(int t=p[i]>>1,las=0;t;t>>=1)Sum=1ll*Sum*t%Mod,ans=(ans+1ll*Sum*(tag[t<<1]+tag[t<<1|1]-las)%Mod*iS[t])%Mod,las=tag[t];
	}Rep(i,l,mid)Del(i,S[i]);
	Solve(x<<1|1,mid+1,r);
	Rep(i,l,mid)Ins(i,S[i]);
}
int main(){
//	freopen("std.in","r",stdin);
//	freopen("std.out","w",stdout);
	H=read();iS[0]=S[0]=1;
	Rep(i,1,(1<<H)-1)S[i]=1ll*S[i>>1]*i%Mod,iS[i]=Fp(S[i],Mod-2);
	Rep(i,(1<<H-1),(1<<H)-1)p[i]=read()+(1<<H-1)-1;
	Solve(1,1<<H-1,(1<<H)-1),cout<<ans<<'\n';
	return 0;
}


