Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.

Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.

Input

The first line contains integer q (1 ≤ q ≤ 1000), the number of handle change requests.

Next q lines contain the descriptions of the requests, one per line.

Each query consists of two non-empty strings old and new, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings old and new are distinct. The lengths of the strings do not exceed 20.

The requests are given chronologically. In other words, by the moment of a query there is a single person with handle old, and handle new is not used and has not been used by anyone.

Output

In the first line output the integer n — the number of users that changed their handles at least once.

In the next n lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, old and new, separated by a space, meaning that before the user had handle old, and after all the requests are completed, his handle is new. You may output lines in any order.

Each user who changes the handle must occur exactly once in this description.

Examples

Input

5
Misha ILoveCodeforces
Vasya Petrov
Petrov VasyaPetrov123
ILoveCodeforces MikeMirzayanov
Petya Ivanov


Output

3
Petya Ivanov
Misha MikeMirzayanov
Vasya VasyaPetrov123
#include<bits/stdc++.h> /*
#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp> // Including */
using namespace std;
//using namespace __gnu_pbds;

//typedefs

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef pair<int,int> pii;
typedef pair<double, double> pdd;
typedef pair<ll, ll> pll;
typedef vector<pii> vii;
typedef vector<pll> vll;
typedef vector<int>::iterator vit;
typedef set<int>::iterator sit;

/*
template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename F, typename S>
using ordered_map = tree<F, S, less<F>, rb_tree_tag, tree_order_statistics_node_update>;
*/

//#Defines
#define rep(i,a,b)	for(i=a;i<=b;i++)
#define repR(i,a,b)	for(i=a;i>=b;i--)
//#define pb push_back
#define pb emplace_back
#define F first
#define S second
#define mp make_pair
#define all(c)		c.begin(),c.end()
#define endl '\n'
#define pf printf
#define sf scanf
//#define left __left
//#define right __right
//#define tree __tree
#define MOD 1000000007
//#define harmonic(n) 0.57721566490153286l+log(n)

#define RESET(a,b)	memset(a,b,sizeof(a))
#define gcd(a,b) __gcd(a,b)
#define lcm(a,b) (a*(b/gcd(a,b)))
#define sqr(a) ((a) * (a))

#define optimize() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define fraction() cout.unsetf(ios::floatfield); cout.precision(10); cout.setf(ios::fixed,ios::floatfield);

const double PI = acos(-1);
const double eps = 1e-9;
const int inf = 2000000000;
const ll infLL = 9000000000000000000;


//Bit Operations
inline bool checkBit(ll n, int i) { return n&(1LL<<i); }
inline ll setBit(ll n, int i) { return n|(1LL<<i);; }
inline ll resetBit(ll n, int i) { return n&(~(1LL<<i)); }

int fx[] = {0, 0, +1, -1};
int fy[] = {+1, -1, 0, 0};
//int dx[] = {+1, 0, -1, 0, +1, +1, -1, -1};
//int dy[] = {0, +1, 0, -1, +1, -1, +1, -1};


//Inline functions

inline bool EQ(double a, double b) { return fabs(a-b) < 1e-9; }
inline bool isLeapYear(ll year) { return (year%400==0) || (year%4==0 && year%100!=0); }
inline void normal(ll &a) { a %= MOD; (a < 0) && (a += MOD); }
inline ll modMul(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); return (a*b)%MOD; }
inline ll modAdd(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); return (a+b)%MOD; }
inline ll modSub(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); a -= b; normal(a); return a; }
inline ll modPow(ll b, ll p) { ll r = 1; while(p) { if(p&1) r = modMul(r, b); b = modMul(b, b); p >>= 1; } return r; }
inline ll modInverse(ll a) { return modPow(a, MOD-2); }
inline ll modDiv(ll a, ll b) { return modMul(a, modInverse(b)); }
inline bool isInside(pii p,ll n,ll m)	{ return (p.first>=0&&p.first<n&&p.second>=0&&p.second<m); }
inline bool isInside(pii p,ll n)		{ return (p.first>=0&&p.first<n&&p.second>=0&&p.second<n); }
inline bool isSquare(ll x) 				{ ll s = sqrt(x);	return (s*s==x); }
inline bool isFib(ll x) 				{ return isSquare(5*x*x+4)|| isSquare(5*x*x-4); }
inline bool isPowerOfTwo(ll x)			{ return ((1LL<<(ll)log2(x))==x); }


struct func
{
	//this is a sample overloading function for sorting stl
	bool operator()(pii const &a, pii const &b)	
	{
		if(a.F==b.F)
			return (a.S<b.S);
		return (a.F<b.F);
	}
};

//Prime Number Generator

/*
#define M 100000000
int marked[M/64 + 2];

#define on(x) (marked[x/64] & (1<<((x%64)/2)))
#define mark(x)  marked[x/64] |= (1<<((x%64)/2))

vl prime;

bool isPrime(int num) {
  return num > 1 && (num == 2 || ((num & 1) && !on(num))); 
}

void sieve(ll n) {
	for (ll i = 3; i * i < n; i += 2) {
		if (!on(i)) {
			for (ll j = i * i; j <= n; j += i + i) {
				mark(j);
			}
		}
	}
	prime.pb(2);
	for(ll i = 3; i <= n; i += 2) {
		if(!on(i))	prime.pb(i);	
	}
}

*/

//
//debug
#ifdef tajir
template < typename F, typename S >
ostream& operator << ( ostream& os, const pair< F, S > & p ) {
	return os << "(" << p.first << ", " << p.second << ")";
}

template < typename T >
ostream &operator << ( ostream & os, const vector< T > &v ) {
	os << "{";
	for(auto it = v.begin(); it != v.end(); ++it) {
		if( it != v.begin() ) os << ", ";
			os << *it;
	}
	return os << "}";
}

template < typename T >
ostream &operator << ( ostream & os, const set< T > &v ) {
	os << "[";
	for(auto it = v.begin(); it != v.end(); ++it) {
		if( it != v.begin() ) os << ", ";
			os << *it;
	}
	return os << "]";
}

template < typename T >
ostream &operator << ( ostream & os, const multiset< T > &v ) {
	os << "[";
	for(auto it = v.begin(); it != v.end(); ++it) {
		if( it != v.begin() ) os << ", ";
		os << *it;
	}
	return os << "]";
}

template < typename F, typename S >
ostream &operator << ( ostream & os, const map< F, S > &v ) {
	os << "[";
	for(auto it = v.begin(); it != v.end(); ++it) {
		if( it != v.begin() ) os << ", ";
		os << it -> first << " = " << it -> second ;
		}
	return os << "]";
}

#define dbg(args...) do {cerr << #args << " : "; faltu(args); } while(0)

clock_t tStart = clock();
#define timeStamp dbg("Execution Time: ", (double)(clock() - tStart)/CLOCKS_PER_SEC)

void faltu () {
	cerr << endl;
}

template <typename T>
void faltu( T a[], int n ) {
	for(int i = 0; i < n; ++i) cerr << a[i] << ' ';
	cerr << endl;
}

template <typename T, typename ... hello>
void faltu( T arg, const hello &... rest) {
	cerr << arg << ' ';
	faltu(rest...);
}
#else
#define dbg(args...)
#endif // tajir
map < string , string > m1;
map < string , string > m2;
int n;

int main()
{
	#ifdef tajir
		freopen("input.txt", "r", stdin);
	#else
		// online submission
	#endif
	optimize();
	int q;
	string a,b;
	cin >> q;
	for(int i = 0; i < q; ++i) {
		cin >> a >> b;
		if(m2.find(a) != m2.end()) {
			m1[m2[a]] = b;
			m2[b] = m2[a];
			m2.erase(a);
			continue;
		}

		m1[a] = b;
		m2[b] = a;
	}
	cout << m1.size() << endl;
	for(auto it = m1.begin(); it != m1.end(); it++)
		cout << it->F << " " << it->S << endl;

	return 0;
}






//?
