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
// And you curse yourself for things you never done

#include<bits/stdc++.h>

#define F first
#define S second
#define PB push_back
#define sz(s) int((s).size())
#define bit(n,k) (((n)>>(k))&1)

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;

const int maxn = 3e5 + 10, mod = 1e9 + 7, inf = 1e9 + 10;

vector<pii> v[maxn];
int cof[maxn], iv[maxn], p[maxn];

int Pow(int a, int b){
    int ans = 1;
    for(; b; b>>=1, a = 1ll * a * a % mod)
	if(b & 1)
	    ans = 1ll * ans * a % mod;
    return ans;
}

int cnt[maxn];

int main(){
    ios_base::sync_with_stdio(false); cin.tie(0); cout.tie();

    iv[0] = iv[1] = 1;
    cof[1] = 1;
    for(int i = 2; i < maxn; i++){
	cof[i] = 1ll * cof[i>>1] * i % mod;
	iv[i] = Pow(cof[i], mod-2);
    }
    
    int n;
    cin >> n;
    --n;
    int L = 1<<n;
    for(int i = 0; i < L; i++){
	cin >> p[i];
	--p[i];
    }
    int ans = 0;
    for(int a = 0; a < n; a++){
	for(int b = 0; b < n; b++){
	    for(int i = 0; i < L; i++){
		int A = L + i, B = L + p[i];
		int num = 1ll * cof[A] * cof[B] % mod;
		v[A>>a].PB({B>>b, num});
	    }
	}
    }
    for(int A = 1; A < maxn - 5; A++){
	if(v[A].empty() || v[A ^ 1].empty())
	    continue;
	for(auto [B, x] : v[A]){
	    cnt[B] = (cnt[B] + x) % mod;
	}
	for(auto [B, x] : v[A ^ 1]){
	    B^=1;
	    int lc1 = A >> 1, lc2 = B >> 1;
	    ans = (1ll * ans + 1ll * cnt[B] * x % mod * iv[lc1] % mod * iv[lc2] % mod * iv[lc1 >> 1] % mod * iv[lc2 >> 1] % mod) % mod;
	}
	for(auto [B, x] : v[A]){
	    cnt[B] = 0;
	}
    }
    ans = 1ll * ans * ((mod + 1) / 2) % mod;
    return cout << ans << endl, 0;
}
