There is a tree T with N vertices, numbered 1 through N. For each 1 ≤ i ≤ N - 1, the i-th edge connects vertices a_i and b_i.

Snuke is constructing a directed graph T' by arbitrarily assigning direction to each edge in T. (There are 2^{N - 1} different ways to construct T'.)

For a fixed T', we will define d(s,\ t) for each 1 ≤ s,\ t ≤ N, as follows:

* d(s,\ t) = (The number of edges that must be traversed against the assigned direction when traveling from vertex s to vertex t)



In particular, d(s,\ s) = 0 for each 1 ≤ s ≤ N. Also note that, in general, d(s,\ t) ≠ d(t,\ s).

We will further define D as the following:

3d2f3f88e8fa23f065c04cd175c14ebf.png

Snuke is constructing T' so that D will be the minimum possible value. How many different ways are there to construct T' so that D will be the minimum possible value, modulo 10^9 + 7?

Constraints

* 2 ≤ N ≤ 1000
* 1 ≤ a_i,\ b_i ≤ N
* The given graph is a tree.

Input

The input is given from Standard Input in the following format:


N
a_1 b_1
a_2 b_2
:
a_{N - 1} b_{N - 1}


Output

Print the number of the different ways to construct T' so that D will be the minimum possible value, modulo 10^9 + 7.

Examples

Input

4
1 2
1 3
1 4


Output

2


Input

4
1 2
2 3
3 4


Output

6


Input

6
1 2
1 3
1 4
2 5
2 6


Output

14


Input

10
2 4
2 5
8 3
10 7
1 6
2 8
9 5
8 6
10 6


Output

102
// ba mei zi !

#include<bits/stdc++.h>

#define HEAP priority_queue
#define rep(i, n) for(int i = 0; i < (n); ++i)
#define per(i, n) for(int i = (n) - 1; i >= 0 ; --i)
#define forn(i, l, r) for(int i = (l); i <= (r); ++i)
#define nrof(i, r, l) for(int i = (r); i >= (l); --i)
#define FOR(a, b) for(auto (a): (b))
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define X first
#define Y second
#define eps 1e-6
#define pi 3.1415926535897932384626433832795
#define SZ(x) (int)x.size()
#define ALL(x) x.begin(), x.end()
#define FILL(a, b) memset((a), (b), sizeof((a)))
#define MCPY(a, b) memcpy((a), (b), sizeof((b)))

using namespace std;

typedef long long LL;
typedef double flt;
typedef vector<int> vi;
typedef vector<LL> vl;
typedef pair<int,int> pii;
typedef pair<int,LL> pil;
typedef pair<LL,int> pli;
typedef pair<LL,LL> pll;
typedef vector<pil> vil;
typedef vector<pii> vii;
typedef vector<pli> vli;
typedef vector<pll> vll;

const int iinf = 1e9 + 7;
const LL linf = 1ll << 60;
const flt dinf = 1e60;

template <typename T>
inline void scf(T &x)
{
	bool f = 0; x = 0; char c = getchar();
	while((c < '0' || c > '9') && c != '-') c = getchar();
	if(c == '-') { f = 1; c = getchar(); }
	while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
	if(f) x = -x; return;
}

template <typename T1, typename T2>
void scf(T1 &x, T2 &y) { scf(x); return scf(y); }

template <typename T1, typename T2, typename T3>
void scf(T1 &x, T2 &y, T3 &z) { scf(x); scf(y); return scf(z); }

template <typename T1, typename T2, typename T3, typename T4>
void scf(T1 &x, T2 &y, T3 &z, T4 &w) { scf(x); scf(y); scf(z); return scf(w); }

inline char mygetchar(){ char c = getchar(); while(c == ' ' || c == '\n') c = getchar(); return c; }

template <typename T>
void chkmax(T &x, const T &y){ if(y > x) x = y; return; }

template <typename T>
void chkmin(T &x, const T &y){ if(y < x) x = y; return; }

#ifdef ONLINE_JUDGE
#define debug(x,c) ;
#else
#define DEBUG
#define debug(x,c) cerr<<#x<<"="<<x<<c;
#endif

//---------------------------head----------------------------

const int N = 1e3 + 100;
const int mod = 1e9 + 7;

int n, diameter_length, min_D;
int dp[N][N / 2], tag[N / 2][N], current_time;
vi g[N];
vi diameter_path;

namespace TREE
{
	int par[N], dep[N];

	void dfs(int u = 1, int fa = 0)
	{
		par[u] = fa;
		dep[u] = dep[fa] + 1;
		for(int v: g[u]) if(v != fa) dfs(v, u);
		return;
	}

	void get_diameter()
	{
		dfs();
		int mx = -1, S, T;
		forn(i, 1, n) if(dep[i] > mx) mx = dep[i], S = i;
		dfs(S);
		mx = -1;
		forn(i, 1, n) if(dep[i] > mx) mx = dep[i], T = i;
		
		diameter_length = dep[T] - 1;
		min_D = dep[T] >> 1;
		
		int u = T;
		while(u != S) diameter_path.pb(u), u = par[u];
		diameter_path.pb(S);
		return;
	}
}

void MEI()
{
	scf(n);
	rep(i, n - 1)
	{
		int u, v;
		scf(u, v);
		g[u].pb(v);
		g[v].pb(u);
	}
	
	TREE::get_diameter();
	return;
}

namespace Bamboo
{
	void solve()
	{
		dp[0][0] = 1;
		forn(i, 1, n - 1)
		{
			dp[i][0] = dp[i][i] = 1;
			forn(j, 1, i - 1) dp[i][j] = (dp[i - 1][j - 1] + dp[i - 1][j]) % mod;
		}
		printf("%d\n", dp[n - 1][min_D] * 2 % mod);
		return;
	}
}

namespace Odd
{
	int root1, root2, only_down, only_up;

	void DP1(int u, int par, int max_down, int max_up)
	{
		if(tag[u][max_down] == current_time) return;

		tag[u][max_down] = current_time;
		int &t = dp[u][max_down];
		t = 1;
		
		for(int v: g[u]) if(v != par && v != root2)
		{
			int foo = 0;
			if(max_down) DP1(v, u, max_down - 1, max_up), foo += dp[v][max_down - 1];
			if(max_up) DP1(v, u, max_down, max_up -  1), foo += dp[v][max_down];
			t = 1ll * t * foo % mod;
		}
		
		if(u == root1 && max_down)
		{
			DP1(root2, root1, max_down, max_up);
			t = 1ll * t * dp[root2][max_down] % mod;
		}
		return;
	}

	void DP2(int u, int par, int max_down, int max_up)
	{
		if(tag[u][max_down] == current_time) return;

		tag[u][max_down] = current_time;
		int &t = dp[u][max_down];
		t = 1;

		for(int v: g[u]) if(v != par)
		{
			int foo = 0;
			if(max_down && v != only_up) DP1(v, u, max_down - 1, max_up), foo += dp[v][max_down - 1];
			if(max_up && v != only_down) DP1(v, u, max_down, max_up - 1), foo += dp[v][max_down];
			t = 1ll * t * foo % mod;
		}
		return;
	}
	
	void solve()
	{
		int ans = 0;
		root1 = diameter_path[min_D], root2 = diameter_path[min_D - 1];
		
		rep(max_down, min_D)
		{
			current_time++;
			DP1(root1, 0, max_down, min_D - 1 - max_down);
			ans += dp[root1][max_down];
			if(ans >= mod) ans -= mod;
		}

		g[diameter_path.back()].pb(n + 1);
		only_up = n + 1; only_down = root2;
		forn(max_down, 0, min_D)
		{
			current_time++;
			DP2(root1, 0, max_down, min_D - max_down);
			ans += dp[root1][max_down];
			if(ans >= mod) ans -= mod;
		}
		
		printf("%d\n", ans * 2 % mod);
		return;
	}
}

namespace Even
{
	void DP(int u, int par, int max_down, int max_up)
	{
		if(tag[u][max_down] == current_time) return;

		tag[u][max_down] = current_time;
		int &t = dp[u][max_down];
		t = 1;

		for(int v: g[u]) if(v != par)
		{
			int foo = 0;
			if(max_down) DP(v, u, max_down - 1, max_up), foo += dp[v][max_down - 1];
			if(max_up) DP(v, u, max_down, max_up - 1), foo += dp[v][max_down];
			t = 1ll * t * foo % mod;
		}
		return;
	}

	void solve()
	{
		int root = diameter_path[min_D];
		int ans = 0;
		forn(max_down, 0, min_D)
		{
			current_time++;
			DP(root, 0, max_down, min_D - max_down);
			ans += dp[root][max_down];
			if(ans >= mod) ans -= mod;
		}
		printf("%d\n", ans);
		return;
	}
}

void ZI()
{
	if(diameter_length == n - 1) Bamboo::solve();
	else if(diameter_length & 1) for(;;), Odd::solve();
	else Even::solve();
	return;
}

/*
Main
code
is
over
.
If
you
are
in
a
hurry
trying
to
hack
me
you
don't
need
to
read
the
part
left
.
Trust
me
.
It
is
useless
.
My
code
always
gets
a
failed
system
test
,
so
it
will
be
pretty
easy
for
you
to
hack
me
.
Just
find
a
test
case
and
hack
.
Wish
you
all
happy
hack
.
 _____
/     \
| . . |
|     |
|  |  |
| --- |
| \_/ |
\_____/

*/

#ifdef DEBUG

struct I{ int x; I(){} I(int y):x(y){} bool operator <(const I &a) const{ return x < a.x; }};

pair<I,LL> time_limit_exceeded;
const bool FLAG = 1;
static vector<I> wrong_answer;
I *RP=new I;

I FUCK=min(I{*RP}, I{0});
static int memory_limit_exceeded[10];
vi O(){cout<<FUCK.x<<endl; vi ert; int x=0;
x=(int)hypot(3.0, 4.0); ert.pb(x);
if(x&1)exit(0);
auto t=lower_bound(ALL(ert), 5);
static list<int> runtime_error_;
return ert;}
#endif // DEBUG

#define MEI int
#define ZI main
MEI ZI()
{
#undef MEI
#undef ZI

    MEI();
    ZI();

#define BA return
#define MEI 0
#define ZI ;
BA MEI ZI
#undef BA
#undef MEI
#undef ZI
}