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;

int H, L;
int p[1 << 19];
const int MOD = 1e9 + 7;
typedef long long LL;
int val[1 << 19];

template<typename T>
void upd(T &x,T y) {
	x += y;
	if (x >= MOD) x -= MOD;
}

vector<int>fuck;

void clear() {
	for (auto t:fuck) {
		val[t] = 0;
	}
	fuck.clear();
}

LL ans;

void solve(int now, LL fap, int flag) {
	//printf("solve %d %lld %d\n",now,fap,flag);
	fap = fap * now % MOD;
	if (now >= L) {
		int idx = now - L + 1;
		int nxt = now + p[idx] - idx;
		//printf("nxt = %d\n",nxt);
		if (flag == 1) {
			for (int i=nxt;i>=1;i>>=1) {
				fap = fap * i % MOD;
				upd(val[i], int(fap));
				fuck.push_back(i);
			}
		} else {
			assert(val[nxt] == 0);
			for (int i=nxt;i>1;i>>=1) {
				fap = fap * i % MOD;
				//printf("%d %lld %lld\n", i, fap, val[i] * fap % MOD);
				upd(ans, 1LL * val[i ^ 1] * (i / 2) % MOD * fap % MOD);
			}
		}
	} else {
		solve(now * 2, fap, flag);
		solve(now * 2 + 1, fap, flag);
	}
}

int main() {
	scanf("%d",&H);
	L = 1 << (H - 1);
	for (int i=1;i<=L;i++) {
		scanf("%d",&p[i]);
	}
	ans = 0;
	for (int LCA=1;LCA<L;LCA++) {
	//for (int LCA=1;LCA<=1;LCA++) {
		solve(LCA * 2, LCA, 1);
		solve(LCA * 2 + 1, 1, 2);
		//printf("!!!! %d %lld\n",LCA,ans);
		clear();
	}
	printf("%lld\n",ans);
}