Lee tried so hard to make a good div.2 D problem to balance his recent contest, but it still doesn't feel good at all. Lee invented it so tediously slow that he managed to develop a phobia about div.2 D problem setting instead. And now he is hiding behind the bushes...

Let's define a Rooted Dead Bush (RDB) of level n as a rooted tree constructed as described below.

A rooted dead bush of level 1 is a single vertex. To construct an RDB of level i we, at first, construct an RDB of level i-1, then for each vertex u: 

  * if u has no children then we will add a single child to it; 
  * if u has one child then we will add two children to it; 
  * if u has more than one child, then we will skip it. 

<image> Rooted Dead Bushes of level 1, 2 and 3.

Let's define a claw as a rooted tree with four vertices: one root vertex (called also as center) with three children. It looks like a claw:

<image> The center of the claw is the vertex with label 1.

Lee has a Rooted Dead Bush of level n. Initially, all vertices of his RDB are green.

In one move, he can choose a claw in his RDB, if all vertices in the claw are green and all vertices of the claw are children of its center, then he colors the claw's vertices in yellow.

He'd like to know the maximum number of yellow vertices he can achieve. Since the answer might be very large, print it modulo 10^9+7.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.

Next t lines contain test cases — one per line.

The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^6) — the level of Lee's RDB.

Output

For each test case, print a single integer — the maximum number of yellow vertices Lee can make modulo 10^9 + 7.

Example

Input


7
1
2
3
4
5
100
2000000


Output


0
0
4
4
12
990998587
804665184

Note

It's easy to see that the answer for RDB of level 1 or 2 is 0.

The answer for RDB of level 3 is 4 since there is only one claw we can choose: \{1, 2, 3, 4\}.

The answer for RDB of level 4 is 4 since we can choose either single claw \{1, 3, 2, 4\} or single claw \{2, 7, 5, 6\}. There are no other claws in the RDB of level 4 (for example, we can't choose \{2, 1, 7, 6\}, since 1 is not a child of center vertex 2).

<image> Rooted Dead Bush of level 4.
3
1234567
1268501
1268499

788765312
999997375
999999350

/*
author: madhav_1999 aka orthodoxparadox
23 June 2020 at 8:11 PM
*/

#include <bits/stdc++.h>
#include <ext/pb_ds/detail/standard_policies.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

#define dbl long double
#define int ll
#define ll long long
#define pii pair<int, int>
#define len(x) (int) x.size()
#define rev(a) reverse(all(a))
#define oiint ostream_iterator<int>(cout, " ") 
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define endl '\n'
#define all(a) a.begin(), a.end()
#define initialise(a, x) memset(a, x, sizeof(a))
#define onlyunique(v) v.erase(unique(all(v)), v.end()); //only for sorted vector
#define inf 2e18
#pragma GCC optimize ("O3")
#pragma GCC optimize ("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define _CRT_SECURE_NO_WARNINGS
#ifdef MADHAV
#define dbg(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){cerr << name << " : " << arg1 << endl;}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
	const char* comma = strchr(names + 1, ',');
	cerr.write(names, comma - names) << " : " << arg1 << "  ";
	__f(comma + 1, args...);}
#else
#define dbg(...) 42
#endif
template <typename X> ostream& operator << (ostream& x,const vector<X>& v){for(int i=0;i<(int)v.size();++i) x<<v[i]<<" ";return x;} 
template <typename X> ostream& operator << (ostream& x,const set<X>& v){for(auto it:v) x<<it<<" ";return x;} 
template <typename X> ostream& operator << (ostream& x,const multiset<X>& v){for(auto it:v) x<<it<<" ";return x;} 
template <typename X, typename Y> ostream& operator << (ostream& x, const pair<X,Y>& v) {x<<v.ff<<" "<<v.ss;return x;} 
template <typename T, typename S> ostream& operator << (ostream& os, const map<T, S>& v) { for (auto it : v) os << it.first << "=>" << it.second << endl; return os; }
typedef tree <int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
struct pair_hash {inline std::size_t operator()(const std::pair<int, int> & v) const {return v.first*31+v.second;}};
const int MOD = 1e9 + 7;

int takemod(int a, int mod = MOD){a%=mod;if(a<0)a+=mod;return a;}

int fast_exp(int base, int expo, int mod = MOD) {
	int res=1;
	while(expo>0) {
		if(expo&1) res=(res*base)%mod;
		base=(base*base)%mod;
		expo>>=1;}
	return res;}

int modinv(int a, int mod = MOD){return takemod(fast_exp(takemod(a, mod), mod-2, mod), mod);}

const int N = 2e6 + 10;
int ans[N + 2];
int dp[N][2];

void precompute() {
	dp[1][0] = 1;
	dp[1][1] = 0;
	ans[1] = ans[2] = 0;
	for(int i = 2; i < N; i++) {
		dp[i][0] = takemod(dp[i - 1][0] + 2*dp[i - 1][1]);
		dp[i][1] = dp[i - 1][0];
		ans[i + 1] = takemod(4 * dp[i][1]);
	}
}


void solve()
{
	int n;
	cin >> n;
	cout << ans[n] << endl;	
}

signed main(){
	ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
	#ifdef MADHAV
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	freopen("error.txt", "w", stderr);
	#endif
	clock_t t1, t2;
	t1 = clock();

	int tt = 1;
	cin >> tt;
	precompute();
	while(tt--)
	{
		solve();
	}
	#ifdef MADHAV
	t2 = clock(); cout << "\n\ntime taken: " << (t2-t1)/(dbl)CLOCKS_PER_SEC;
	#endif
	return 0;
}