Problem: We have a tree G with N vertices numbered 1 to N. The i-th edge of G connects Vertex a_i and Vertex b_i.

Consider adding zero or more edges in G, and let H be the graph resulted.

Find the number of graphs H that satisfy the following conditions, modulo 998244353.

* H does not contain self-loops or multiple edges.
* The diameters of G and H are equal.
* For every pair of vertices in H that is not directly connected by an edge, the addition of an edge directly connecting them would reduce the diameter of the graph.

Constraints

* 3 \le N \le 2 \times 10^5
* 1 \le a_i, b_i \le N
* The given graph is a tree.

Input

Input is given from Standard Input in the following format:


N
a_1 b_1
\vdots
a_{N-1} b_{N-1}


Output

Print the answer.

Examples

Input

6
1 6
2 1
5 2
3 4
2 3


Output

3


Input

3
1 2
2 3


Output

1


Input

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


Output

27


Input

19
2 4
15 8
1 16
1 3
12 19
1 18
7 11
11 15
12 9
1 6
7 14
18 2
13 12
13 5
16 13
7 1
11 10
7 17


Output

78732

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
 
typedef long long ll;
typedef long double ld;
typedef double db; 
typedef string str; 

typedef pair<int,int> pi;
typedef pair<ll,ll> pl; 
typedef pair<db,db> pd; 

typedef vector<int> vi; 
typedef vector<ll> vl; 
typedef vector<db> vd; 
typedef vector<str> vs; 
typedef vector<pi> vpi;
typedef vector<pl> vpl; 
typedef vector<pd> vpd; 

#define mp make_pair 
#define f first
#define s second
#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define rall(x) (x).rbegin(), (x).rend() 
#define rsz resize
#define ins insert 
#define ft front() 
#define bk back()
#define pf push_front 
#define pb push_back
#define eb emplace_back 
#define lb lower_bound 
#define ub upper_bound 

#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)

const int MOD = 998244353;
const int MX = 2e5+5; 
const ll INF = 1e18; 
const ld PI = acos((ld)-1);
const int xd[4] = {1,0,-1,0}, yd[4] = {0,1,0,-1}; 

template<class T> bool ckmin(T& a, const T& b) { 
	return b < a ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) { 
	return a < b ? a = b, 1 : 0; } 
int pct(int x) { return __builtin_popcount(x); } 
int bit(int x) { return 31-__builtin_clz(x); } // floor(log2(x)) 
int cdiv(int a, int b) { return a/b+!(a<0||a%b == 0); } // division of a by b rounded up, assumes b > 0 

namespace input {
	template<class T> void re(complex<T>& x);
	template<class T1, class T2> void re(pair<T1,T2>& p);
	template<class T> void re(vector<T>& a);
	template<class T, size_t SZ> void re(array<T,SZ>& a);

	template<class T> void re(T& x) { cin >> x; }
	void re(double& x) { string t; re(t); x = stod(t); }
	void re(ld& x) { string t; re(t); x = stold(t); }
	template<class T, class... Ts> void re(T& t, Ts&... ts) { 
		re(t); re(ts...); 
	}

	template<class T> void re(complex<T>& x) { T a,b; re(a,b); x = {a,b}; }
	template<class T1, class T2> void re(pair<T1,T2>& p) { re(p.f,p.s); }
	template<class T> void re(vector<T>& a) { F0R(i,sz(a)) re(a[i]); }
	template<class T, size_t SZ> void re(array<T,SZ>& a) { F0R(i,SZ) re(a[i]); }
}

using namespace input;

namespace output {
	void pr(int x) { cout << x; }
	void pr(long x) { cout << x; }
	void pr(ll x) { cout << x; }
	void pr(unsigned x) { cout << x; }
	void pr(unsigned long x) { cout << x; }
	void pr(unsigned long long x) { cout << x; }
	void pr(float x) { cout << x; }
	void pr(double x) { cout << x; }
	void pr(ld x) { cout << x; }
	void pr(char x) { cout << x; }
	void pr(const char* x) { cout << x; }
	void pr(const string& x) { cout << x; }
	void pr(bool x) { pr(x ? "true" : "false"); }
	template<class T> void pr(const complex<T>& x) { cout << x; }
	
	template<class T1, class T2> void pr(const pair<T1,T2>& x);
	template<class T> void pr(const T& x);
	
	template<class T, class... Ts> void pr(const T& t, const Ts&... ts) { 
		pr(t); pr(ts...); 
	}
	template<class T1, class T2> void pr(const pair<T1,T2>& x) { 
		pr("{",x.f,", ",x.s,"}"); 
	}
	template<class T> void pr(const T& x) { 
		pr("{"); // const iterator needed for vector<bool>
		bool fst = 1; for (const auto& a: x) pr(!fst?", ":"",a), fst = 0; 
		pr("}");
	}
	
	void ps() { pr("\n"); } // print w/ spaces
	template<class T, class... Ts> void ps(const T& t, const Ts&... ts) { 
		pr(t); if (sizeof...(ts)) pr(" "); ps(ts...); 
	}
	
	void pc() { pr("]\n"); } // debug w/ commas
	template<class T, class... Ts> void pc(const T& t, const Ts&... ts) { 
		pr(t); if (sizeof...(ts)) pr(", "); pc(ts...); 
	}
	#define dbg(x...) pr("[",#x,"] = ["), pc(x);
}

using namespace output;

namespace io {
	void setIn(string s) { freopen(s.c_str(),"r",stdin); }
	void setOut(string s) { freopen(s.c_str(),"w",stdout); }
	void setIO(string s = "") {
		ios_base::sync_with_stdio(0); cin.tie(0); // fast I/O
		// cin.exceptions(cin.failbit); 
		// throws exception when do smth illegal
		// ex. try to read letter into int
		if (sz(s)) { setIn(s+".in"), setOut(s+".out"); } // for USACO
	}
}

using namespace io;

mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count()); 

/**
 * Description: modular arithmetic operations 
 * Source: 
	* KACTL
	* https://codeforces.com/blog/entry/63903
	* https://codeforces.com/contest/1261/submission/65632855 (tourist)
	* https://codeforces.com/contest/1264/submission/66344993 (ksun)
 * Verification: 
	* https://open.kattis.com/problems/modulararithmetic
 */

struct mi {
	typedef decay<decltype(MOD)>::type T; 
 	/// don't silently convert to T
	T v; explicit operator T() const { return v; }
	mi() { v = 0; }
	mi(ll _v) { 
		v = (-MOD < _v && _v < MOD) ? _v : _v % MOD;
		if (v < 0) v += MOD;
	}
	friend bool operator==(const mi& a, const mi& b) { 
		return a.v == b.v; }
	friend bool operator!=(const mi& a, const mi& b) { 
		return !(a == b); }
	friend bool operator<(const mi& a, const mi& b) { 
		return a.v < b.v; }
	friend void re(mi& a) { ll x; re(x); a = mi(x); }
	friend void pr(const mi& a) { pr(a.v); }
	friend ostream& operator<<(ostream& os, const mi& a) { 
		return os << a.v; }
   
	mi& operator+=(const mi& m) { 
		if ((v += m.v) >= MOD) v -= MOD; 
		return *this; }
	mi& operator-=(const mi& m) { 
		if ((v -= m.v) < 0) v += MOD; 
		return *this; }
	mi& operator*=(const mi& m) { 
		v = (ll)v*m.v%MOD; return *this; }
	mi& operator/=(const mi& m) { return (*this) *= inv(m); }
	friend mi pow(mi a, ll p) {
		mi ans = 1; assert(p >= 0);
		for (; p; p /= 2, a *= a) if (p&1) ans *= a;
		return ans;
	}
	friend mi inv(const mi& a) { assert(a.v != 0); 
		return pow(a,MOD-2); }
		
	mi operator-() const { return mi(-v); }
	mi& operator++() { return *this += 1; }
	mi& operator--() { return *this -= 1; }
	friend mi operator+(mi a, const mi& b) { return a += b; }
	friend mi operator-(mi a, const mi& b) { return a -= b; }
	friend mi operator*(mi a, const mi& b) { return a *= b; }
	friend mi operator/(mi a, const mi& b) { return a /= b; }
};
typedef vector<mi> vmi;
typedef pair<mi,mi> pmi;
typedef vector<pmi> vpmi;

vector<vmi> comb;
void genComb(int SZ) {
	comb.assign(SZ,vmi(SZ)); comb[0][0] = 1;
	FOR(i,1,SZ) F0R(j,i+1) 
		comb[i][j] = comb[i-1][j]+(j?comb[i-1][j-1]:0);
}

/**
 * Description: Calculates longest path in tree
 * Source: own
 * Verification: 
   * http://www.spoj.com/problems/PT07Z/
   * https://codeforces.com/contest/1182/problem/D
 */

template<int SZ> struct TreeDiameter {
	int N; 
	vi adj[SZ];
	void ae(int a, int b) { adj[a].pb(b), adj[b].pb(a); }
	int par[SZ], dist[SZ];
	void dfs(int x) {
		trav(y,adj[x]) if (y != par[x]) {
			par[y] = x; dist[y] = dist[x]+1;
			dfs(y);
		}
	}
	void genDist(int x) { par[x] = -1; dist[x] = 0; dfs(x); }
	int diaLen;
	vi center, dia = {1,1}; 
	void init(int _N) {
		N = _N; // find one endpoint of a diameter
		genDist(1); FOR(i,1,N+1) if (dist[i]>dist[dia[0]]) dia[0] = i; 
		genDist(dia[0]); FOR(i,1,N+1) if (dist[i]>dist[dia[1]]) dia[1] = i;
		diaLen = dist[dia[1]];
		int cen = dia[1]; F0R(i,diaLen/2) cen = par[cen];
		if (diaLen&1) center = {cen,par[cen]};
		else center = {cen};
	}
};
TreeDiameter<MX> T;

mi dp[MX][3];
// not possible -> 0, yes possible -> 0, yes possible -> 1
int N, dist[MX];

void genDist(int a, int b) {
	trav(t,T.adj[a]) if (t != b) {
		dist[t] = dist[a]+1;
		genDist(t,a);
	}
}

void solve(int a, int b) {
	dp[a][0] = 1;
	mi zero = 1, one = 0;
	if (dist[a] == T.diaLen/2) zero = 0, one = 1;
	trav(t,T.adj[a]) if (t != b) {
		solve(t,a);
		dp[a][0] *= 3*dp[t][0];
		mi t0 = 2*dp[t][0]+dp[t][1], t1 = dp[t][2];
		tie(zero,one) = mp(zero*t0,zero*t1+one*t0);
	}
	dp[a][1] = zero; dp[a][2] = one;
}

mi cur[3];

void ad(mi no, mi yes) {
	R0F(i,3) {
		if (i < 2) cur[i+1] += cur[i]*yes;
		cur[i] *= no;
	}
	// ps("AD",no,yes);
}

int main() {
	setIO(); re(N);
	F0R(i,N-1) {
		int a,b; re(a,b);
		T.ae(a,b);
	}
	T.init(N);
	// ps(T.center,T.diaLen);
	if (sz(T.center) == 1) {
		int x = T.center[0];
		genDist(x,0);
		cur[0] = 1;
		trav(t,T.adj[x]) {
			solve(t,x);
			ad(dp[t][0]+2*dp[t][1],dp[t][2]);
		}
		ps(cur[2]);
	} else {
		int x = T.center[0], y = T.center[1];
		genDist(x,y); genDist(y,x);
		solve(x,y); solve(y,x);
		ps(dp[x][2]*dp[y][2]);
	}
	// you should actually read the stuff at the bottom
}

/* stuff you should look for
	* int overflow, array bounds
	* special cases (n=1?)
	* do smth instead of nothing and stay organized
	* WRITE STUFF DOWN
*/