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;

#define pii pair<int, int>

const int maxN = 1 << 18 | 5, mod = 1e9 + 7;

int H, L;
int res[maxN + 1], ans[maxN + 1];
int w[maxN + 1];

vector<pii> vec[maxN + 1];

inline void update(int &x, int y) { x = x + y >= mod ? x + y - mod : x + y; }

int main()
{
	scanf("%d ", &H);
	L = 1 << H - 1;
	for(int i = 1; i <= L; i++)
	{
		int x = L + i - 1, p;
		scanf("%d", &p);
		p = L + p - 1;
		while(x) vec[x].push_back( pii(L + i - 1, p) ), x >>= 1;
		w[L + i - 1] = L + i - 1;
	}
	for(int x = L - 1; x >= 1; x--)
	{
		for(int i = 0; i < vec[x].size(); i++)
		{
			int u = vec[x][i].first, v = vec[x][i].second;
			int tp = 1;
			while(v)
			{
				tp = 1ll * tp * v % mod;
				update(ans[x], 1ll * w[u] * x % mod * res[v] % mod * tp % mod);
				update(ans[x], mod - 1ll * w[u] * x % mod * res[v] % mod * tp % mod * v % mod * (v >> 1) % mod);
				v >>= 1;
			}
			v = vec[x][i].second; tp = 1;
			while(v) 
			{
				update(res[v], 1ll * w[u] * tp % mod);
				tp = 1ll * tp * v % mod;
				v >>= 1;
			}
		}
		for(int i = 0; i < vec[x].size(); i++)
		{
			int u = vec[x][i].first, v = vec[x][i].second;
			while(v) res[v] = 0, v >>= 1;
			w[u] = 1ll * w[u] * x % mod;
		}
	}
	int last = 0;
	for(int x = 1; x < L; x++)
	{
		update(ans[x], mod - 1ll * x * (x << 1) % mod * ans[x << 1] % mod);
		update(ans[x], mod - 1ll * x * (x << 1 | 1) % mod * ans[x << 1 | 1] % mod);
		update(last, ans[x]); 
	}
	printf("%d\n", last);
	return 0;
}
