Nezzar designs a brand new game "Hidden Permutations" and shares it with his best friend, Nanako.

At the beginning of the game, Nanako and Nezzar both know integers n and m. The game goes in the following way:

  * Firstly, Nezzar hides two permutations p_1,p_2,…,p_n and q_1,q_2,…,q_n of integers from 1 to n, and Nanako secretly selects m unordered pairs (l_1,r_1),(l_2,r_2),…,(l_m,r_m); 
  * After that, Nanako sends his chosen pairs to Nezzar; 
  * On receiving those m unordered pairs, Nezzar checks if there exists 1 ≤ i ≤ m, such that (p_{l_i}-p_{r_i}) and (q_{l_i}-q_{r_i}) have different signs. If so, Nezzar instantly loses the game and gets a score of -1. Otherwise, the score Nezzar gets is equal to the number of indices 1 ≤ i ≤ n such that p_i ≠ q_i. 



However, Nezzar accidentally knows Nanako's unordered pairs and decides to take advantage of them. Please help Nezzar find out two permutations p and q such that the score is maximized.

Input

The first line contains a single integer t (1 ≤ t ≤ 5 ⋅ 10^5) — the number of test cases.

The first line of each test case contains two integers n,m (1 ≤ n ≤ 5 ⋅ 10^5, 0 ≤ m ≤ min((n(n-1))/(2),5 ⋅ 10^5)). 

Then m lines follow, i-th of them contains two integers l_i,r_i (1 ≤ l_i,r_i ≤ n, l_i ≠ r_i), describing the i-th unordered pair Nanako chooses. It is guaranteed that all m unordered pairs are distinct.

It is guaranteed that the sum of n for all test cases does not exceed 5 ⋅ 10^5, and the sum of m for all test cases does not exceed 5⋅ 10^5.

Output

For each test case, print two permutations p_1,p_2,…,p_n and q_1,q_2,…,q_n such that the score Nezzar gets is maximized.

Example

Input


3
4 2
1 2
3 4
6 4
1 2
1 3
3 5
3 6
2 1
1 2


Output


1 2 3 4
3 4 1 2
2 3 4 1 6 5
1 4 3 2 5 6
1 2
1 2

Note

For first test case, for each pair given by Nanako:

  * for the first pair (1,2): p_1 - p_2 = 1 - 2 = -1, q_1 - q_2 = 3 - 4 = -1, they have the same sign; 
  * for the second pair (3,4): p_3 - p_4 = 3 - 4 = -1, q_3 - q_4 = 1 - 2 = -1, they have the same sign. 



As Nezzar does not lose instantly, Nezzar gains the score of 4 as p_i ≠ q_i for all 1 ≤ i ≤ 4. Obviously, it is the maximum possible score Nezzar can get.
#include <bits/stdc++.h>
using namespace std;
 
using ll = long long;
using db = long double; // or double, if TL is tight
using str = string; // yay python!

using pi = pair<int,int>;
using pl = pair<ll,ll>;
using pd = pair<db,db>;

using vi = vector<int>;
using vb = vector<bool>;
using vl = vector<ll>;
using vd = vector<db>; 
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>; 
using vpd = vector<pd>;

#define tcT template<class T
#define tcTU tcT, class U
// ^ lol this makes everything look weird but I'll try it
tcT> using V = vector<T>; 
tcT, size_t SZ> using AR = array<T,SZ>; 
tcT> using PR = pair<T,T>;

// pairs
#define mp make_pair
#define f first
#define s second

// vectors
// oops size(x), rbegin(x), rend(x) need C++17
#define sz(x) int((x).size())
#define bg(x) begin(x)
#define all(x) bg(x), end(x)
#define rall(x) x.rbegin(), x.rend() 
#define sor(x) sort(all(x)) 
#define rsz resize
#define ins insert 
#define ft front()
#define bk back()
#define pb push_back
#define eb emplace_back 
#define pf push_front
#define rtn return

#define lb lower_bound
#define ub upper_bound 
tcT> int lwb(V<T>& a, const T& b) { return int(lb(all(a),b)-bg(a)); }

// loops
#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 EACH(a,x) for (auto& a: x)

const int MOD = 1e9+7; // 998244353;
const int MX = 5e5+5;
const ll INF = 1e18; // not too close to LLONG_MAX
const db PI = acos((db)-1);
const int dx[4] = {1,0,-1,0}, dy[4] = {0,1,0,-1}; // for every grid problem!!
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count()); 
template<class T> using pqg = priority_queue<T,vector<T>,greater<T>>;

// bitwise ops
// also see https://gcc.gnu.org/onlinedocs/gcc/Other-Builtins.html
constexpr int pct(int x) { return __builtin_popcount(x); } // # of bits set
constexpr int bits(int x) { // assert(x >= 0); // make C++11 compatible until USACO updates ...
	return x == 0 ? 0 : 31-__builtin_clz(x); } // floor(log2(x)) 
constexpr int p2(int x) { return 1<<x; }
constexpr int msk2(int x) { return p2(x)-1; }

ll cdiv(ll a, ll b) { return a/b+((a^b)>0&&a%b); } // divide a by b rounded up
ll fdiv(ll a, ll b) { return a/b-((a^b)<0&&a%b); } // divide a by b rounded down

tcT> bool ckmin(T& a, const T& b) {
	return b < a ? a = b, 1 : 0; } // set a = min(a,b)
tcT> bool ckmax(T& a, const T& b) {
	return a < b ? a = b, 1 : 0; }

tcTU> T fstTrue(T lo, T hi, U f) {
	hi ++; assert(lo <= hi); // assuming f is increasing
	while (lo < hi) { // find first index such that f is true 
		T mid = lo+(hi-lo)/2;
		f(mid) ? hi = mid : lo = mid+1; 
	} 
	return lo;
}
tcTU> T lstTrue(T lo, T hi, U f) {
	lo --; assert(lo <= hi); // assuming f is decreasing
	while (lo < hi) { // find first index such that f is true 
		T mid = lo+(hi-lo+1)/2;
		f(mid) ? lo = mid : hi = mid-1;
	} 
	return lo;
}
tcT> void remDup(vector<T>& v) { // sort and remove duplicates
	sort(all(v)); v.erase(unique(all(v)),end(v)); }
tcTU> void erase(T& t, const U& u) { // don't erase
	auto it = t.find(u); assert(it != end(t));
	t.erase(it); } // element that doesn't exist from (multi)set

// INPUT
#define tcTUU tcT, class ...U
tcT> void re(complex<T>& c);
tcTU> void re(pair<T,U>& p);
tcT> void re(V<T>& v);
tcT, size_t SZ> void re(AR<T,SZ>& a);

tcT> void re(T& x) { cin >> x; }
void re(double& d) { str t; re(t); d = stod(t); }
void re(long double& d) { str t; re(t); d = stold(t); }
tcTUU> void re(T& t, U&... u) { re(t); re(u...); }

tcT> void re(complex<T>& c) { T a,b; re(a,b); c = {a,b}; }
tcTU> void re(pair<T,U>& p) { re(p.f,p.s); }
tcT> void re(V<T>& x) { EACH(a,x) re(a); }
tcT, size_t SZ> void re(AR<T,SZ>& x) { EACH(a,x) re(a); }
tcT> void rv(int n, V<T>& x) { x.rsz(n); re(x); }

// TO_STRING
#define ts to_string
str ts(char c) { return str(1,c); }
str ts(const char* s) { return (str)s; }
str ts(str s) { return s; }
str ts(bool b) { 
	// #ifdef LOCAL
	// 	return b ? "true" : "false"; 
	// #else 
	return ts((int)b);
	// #endif
}
tcT> str ts(complex<T> c) { 
	stringstream ss; ss << c; return ss.str(); }
str ts(V<bool> v) {
	str res = "{"; F0R(i,sz(v)) res += char('0'+v[i]);
	res += "}"; return res; }
template<size_t SZ> str ts(bitset<SZ> b) {
	str res = ""; F0R(i,SZ) res += char('0'+b[i]);
	return res; }
tcTU> str ts(pair<T,U> p);
tcT> str ts(T v) { // containers with begin(), end()
	#ifdef LOCAL
		bool fst = 1; str res = "{";
		for (const auto& x: v) {
			if (!fst) res += ", ";
			fst = 0; res += ts(x);
		}
		res += "}"; return res;
	#else
		bool fst = 1; str res = "";
		for (const auto& x: v) {
			if (!fst) res += " ";
			fst = 0; res += ts(x);
		}
		return res;

	#endif
}
tcTU> str ts(pair<T,U> p) {
	#ifdef LOCAL
		return "("+ts(p.f)+", "+ts(p.s)+")"; 
	#else
		return ts(p.f)+" "+ts(p.s);
	#endif
}

// OUTPUT
tcT> void pr(T x) { cout << ts(x); }
tcTUU> void pr(const T& t, const U&... u) { 
	pr(t); pr(u...); }
void ps() { pr("\n"); } // print w/ spaces
tcTUU> void ps(const T& t, const U&... u) { 
	pr(t); if (sizeof...(u)) pr(" "); ps(u...); }

// DEBUG
void DBG() { cerr << "]" << endl; }
tcTUU> void DBG(const T& t, const U&... u) {
	cerr << ts(t); if (sizeof...(u)) cerr << ", ";
	DBG(u...); }
#ifdef LOCAL // compile with -DLOCAL, chk -> fake assert
	#define dbg(...) cerr << "Line(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)
	#define chk(...) if (!(__VA_ARGS__)) cerr << "Line(" << __LINE__ << ") -> function(" \
		 << __FUNCTION__  << ") -> CHK FAILED: (" << #__VA_ARGS__ << ")" << "\n", exit(0);
#else
	#define dbg(...) 0
	#define chk(...) 0
#endif

void setPrec() { cout << fixed << setprecision(15); }
void unsyncIO() { cin.tie(0)->sync_with_stdio(0); }
// FILE I/O
void setIn(str s) { freopen(s.c_str(),"r",stdin); }
void setOut(str s) { freopen(s.c_str(),"w",stdout); }
void setIO(str s = "") {
	unsyncIO(); setPrec();
	// 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
}

/**
 * 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)
	* also see https://github.com/ecnerwala/cp-book/blob/master/src/modnum.hpp (ecnerwal)
 * Verification: 
	* https://open.kattis.com/problems/modulararithmetic
 */

template<int MOD, int RT> struct mint {
	static const int mod = MOD;
	static constexpr mint rt() { return RT; } // primitive root for FFT
	int v; explicit operator int() const { return v; } // explicit -> don't silently convert to int
	mint() { v = 0; }
	mint(ll _v) { v = int((-MOD < _v && _v < MOD) ? _v : _v % MOD);
		if (v < 0) v += MOD; }
	friend bool operator==(const mint& a, const mint& b) { 
		return a.v == b.v; }
	friend bool operator!=(const mint& a, const mint& b) { 
		return !(a == b); }
	friend bool operator<(const mint& a, const mint& b) { 
		return a.v < b.v; }
	friend void re(mint& a) { ll x; re(x); a = mint(x); }
	friend str ts(mint a) { return ts(a.v); }
   
	mint& operator+=(const mint& m) { 
		if ((v += m.v) >= MOD) v -= MOD; 
		return *this; }
	mint& operator-=(const mint& m) { 
		if ((v -= m.v) < 0) v += MOD; 
		return *this; }
	mint& operator*=(const mint& m) { 
		v = int((ll)v*m.v%MOD); return *this; }
	mint& operator/=(const mint& m) { return (*this) *= inv(m); }
	friend mint pow(mint a, ll p) {
		mint ans = 1; assert(p >= 0);
		for (; p; p /= 2, a *= a) if (p&1) ans *= a;
		return ans; }
	friend mint inv(const mint& a) { assert(a.v != 0); 
		return pow(a,MOD-2); }
		
	mint operator-() const { return mint(-v); }
	mint& operator++() { return *this += 1; }
	mint& operator--() { return *this -= 1; }
	friend mint operator+(mint a, const mint& b) { return a += b; }
	friend mint operator-(mint a, const mint& b) { return a -= b; }
	friend mint operator*(mint a, const mint& b) { return a *= b; }
	friend mint operator/(mint a, const mint& b) { return a /= b; }
};

typedef mint<MOD,5> mi; // 5 is primitive root for both common mods
typedef vector<mi> vmi;
typedef pair<mi,mi> pmi;
typedef vector<pmi> vpmi;

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

/**
 * Description: pre-compute factorial mod inverses,
 	* assumes $MOD$ is prime and $SZ < MOD$.
 * Time: O(SZ)
 * Source: KACTL
 * Verification: https://dmoj.ca/problem/tle17c4p5
 */

vi invs, fac, ifac; // make sure to convert to LL before doing any multiplications ...
void genFac(int SZ) {
	invs.rsz(SZ), fac.rsz(SZ), ifac.rsz(SZ); 
	invs[1] = fac[0] = ifac[0] = 1; 
	FOR(i,2,SZ) invs[i] = int(MOD-(ll)MOD/i*invs[MOD%i]%MOD);
	FOR(i,1,SZ) {
		fac[i] = int((ll)fac[i-1]*i%MOD);
		ifac[i] = int((ll)ifac[i-1]*invs[i]%MOD);
	}
}
/**
ll comb(int a, int b) {
	if (a < b || b < 0) return 0;
	return (ll)fac[a]*ifac[b]%MOD*ifac[a-b]%MOD;
}
*/

// make sure to intialize ALL GLOBAL VARS between tcs!

// idea: fix directions of edges, find > 1 toposort

int N,M;
set<int> adj[MX];
vi tre[MX];
vpi ed;
vi lef;
vi p,q;

void ae(int x, int y) {
	dbg("AE",x,y);
	tre[x].pb(y);
	tre[y].pb(x);
}

void dfs(int x) {
	vi LEF, cool;
	EACH(t,lef) {
		if (adj[x].count(t)) LEF.pb(t);
		else cool.pb(t);
	}
	swap(lef,LEF);
	EACH(t,cool) {
		ae(x,t);
		dfs(t);
	}
}

vb done;

bool fill_ans(int x) {
	assert(!done[x]); done[x] = 1;
	vi child;
	EACH(t,tre[x]) if (!done[t]) {
		if (!fill_ans(t)) child.pb(t);
	}
	if (!sz(child)) return 0;
	EACH(a,child) p.pb(a);
	p.pb(x);
	q.pb(x);
	EACH(a,child) q.pb(a);
	return 1;
}

void prin(vi v) {
	vi res(N+1);
	int cnt = 0;
	EACH(t,v) {
		res[t] = ++cnt;
	}
	FOR(i,1,N+1) {
		pr(res[i],' ');
	}
	ps();
}

void solve(int tc) {
	re(N,M);
	FOR(i,1,N+1) adj[i].clear(), tre[i].clear();
	F0R(i,M) {
		int l,r; re(l,r);
		adj[l].ins(r), adj[r].ins(l);
	}
	lef.clear();
	FOR(i,1,N+1) lef.pb(i);
	while (sz(lef)) {
		int x = lef.bk; lef.pop_back();
		dfs(x);
	}
	p.clear(), q.clear();
	FOR(i,1,N+1) if (!sz(tre[i])) {
		p.pb(i);
		q.pb(i);
	}
	done = vb(N+1);
	FOR(i,1,N+1) if (sz(tre[i]) == 1 && !done[i]) assert(fill_ans(tre[i][0]));
	assert(sz(p) == N && sz(q) == N);
	prin(p);
	prin(q);
	// EACH(t,p) pr(t,' ');
	// ps();
	// EACH(t,q) pr(t,' ');
	// ps();
	// done = vb(N+1);
	// FOR(i,1,N+1) if (!done[i])

	// vb vis(N+1);
	// FOR(i,1,N+1) if (!vis[i]) {
	// 	dfs(i);
	// }
}

int main() {
	setIO();
	int TC; re(TC);
	FOR(i,1,TC+1) solve(i);
}

/* stuff you should look for
	* int overflow, array bounds
	* special cases (n=1?)
	* do smth instead of nothing and stay organized
	* WRITE STUFF DOWN
	* DON'T GET STUCK ON ONE APPROACH
*/
