Problem: The King of Berland Polycarp LXXXIV has n daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are n other kingdoms as well.

So Polycarp LXXXIV has enumerated his daughters from 1 to n and the kingdoms from 1 to n. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.

Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.

For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the n-th daughter.

For example, let there be 4 daughters and kingdoms, the lists daughters have are [2, 3], [1, 2], [3, 4], [3], respectively.

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In that case daughter 1 marries the prince of kingdom 2, daughter 2 marries the prince of kingdom 1, daughter 3 marries the prince of kingdom 3, leaving daughter 4 nobody to marry to.

Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.

Polycarp LXXXIV wants to increase the number of married couples.

Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.

If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.

For your and our convenience you are asked to answer t independent test cases.

Input

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

Then t test cases follow.

The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of daughters and the number of kingdoms.

Each of the next n lines contains the description of each daughter's list. The first integer k (0 ≤ k ≤ n) is the number of entries in the i-th daughter's list. After that k distinct integers follow g_i[1], g_i[2], ..., g_i[k] (1 ≤ g_i[j] ≤ n) — the indices of the kingdoms in the list in the increasing order (g_i[1] < g_i[2] < ... < g_i[k]).

It's guaranteed that the total number of daughters over all test cases does not exceed 10^5.

It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed 10^5.

Output

For each test case print the answer to it.

Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list.

If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.

Otherwise the only line should contain one word "OPTIMAL".

Example

Input


5
4
2 2 3
2 1 2
2 3 4
1 3
2
0
0
3
3 1 2 3
3 1 2 3
3 1 2 3
1
1 1
4
1 1
1 2
1 3
1 4


Output


IMPROVE
4 4
IMPROVE
1 1
OPTIMAL
OPTIMAL
OPTIMAL

Note

The first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.

In the second test case any new entry will increase the number of marriages from 0 to 1.

In the third and the fourth test cases there is no way to add an entry.

In the fifth test case there is no way to change the marriages by adding any entry.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.BitSet;
import java.util.HashMap;
import java.util.HashSet;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class Main {
	static ArrayList<Integer> adj[];
	static PrintWriter out = new PrintWriter(System.out);
	public static long mod;

	static long[][] notmemo;
	static int k;
	static int a[];
	static int b[];
	static int m;
	static char c[];


	static int trace[];
	static int h;
	static int x;
	static int ans1;
	static int ans2;
    static char t[];
	static char l[];
	static char r[];
    static int w;
    static int s;
	public static void main(String args[]) throws IOException {
		Scanner sc = new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0) {
		int n=sc.nextInt();
		HashSet<Integer> set=new HashSet<Integer>();
		int idx=-1;
		
		for(int i=0 ; i<n ; i++) {
			int k=sc.nextInt();
			boolean flag=false;
			while(k-->0) {
			int x=sc.nextInt();
			if(!set.contains(x)&&!flag) {
				set.add(x);
				flag=true;
			}
			
		}
			if(!flag&&idx==-1) {
				idx=i;
			}
		}
		boolean f2=true;
		if(idx!=-1)
		for (int i = 1; i <=n; i++) {
			if(!set.contains(i)) {
				out.println("IMPROVE");
				out.println((idx+1)+" "+i);
			f2=false;
			break;
			}
		}
		if(f2) {
			out.println("OPTIMAL");
		}
		}
		
		
		out.flush();
    }
    static long dp(int i,int sum) {
		if(i==n) {
			if(sum==s)
			return 1;
			else
				return 0;
		}
		if(notmemo[i][sum]!=-1) {
			return notmemo[i][sum];
		}
		long total=0;
	if(sum==s) {
total=(1+dp(i+1,sum)%998244353)%998244353;
	}	
	else if (sum + a[i] <= s) {
		total=((total%998244353)+dp(i+1,sum+a[i])%998244353)%998244353;
	}
	if(sum!=s)
	total=((total%998244353)+dp(i+1,sum)%998244353)%998244353;
	
	return notmemo[i][sum]=total;
		
	
    }
    static String y;
    	static int nomnom[];
	static long fac[];
	
	static class book implements Comparable<book> {
		int idx;
		long score;

		public book(int i, long s) {
			idx = i;
			score = s;
		}

		public int compareTo(book o) {
			return (int) (o.score - score);
		}
	}

	static class library implements Comparable<library> {
		int numofbooks;
		int signup;
		int shiprate;
		int idx;

		public library(int a, int b, int c, int idx) {
			numofbooks = a;
			signup = b;
			shiprate = c;
			this.idx = idx;
		}

		@Override
		public int compareTo(library o) {
		if(signup==o.signup) {
			  return o.numofbooks-numofbooks;
		}
		return signup - o.signup;
		
		}
	}





	static boolean isOn(int S, int j) {
		return (S & 1 << j) != 0;
	}


	static boolean f = true;

	static class SegmentTree { // 1-based DS, OOP

		int N; // the number of elements in the array as a power of 2 (i.e. after padding)
		int[] array, sTree, lazy;

		SegmentTree(int[] in) {
			array = in;
			N = in.length - 1;
			sTree = new int[N << 1]; // no. of nodes = 2*N - 1, we add one to cross out index zero
			lazy = new int[N << 1];
			//build(1, 1, N);
		}

		void build(int node, int b, int e) // O(n)
		{
			if (b == e)
				sTree[node] = array[b];
			else {
				int mid = b + e >> 1;
				build(node << 1, b, mid);
				build(node << 1 | 1, mid + 1, e);
				sTree[node] = sTree[node << 1] + sTree[node << 1 | 1];
			}
		}

		void update_point(int index, int val) // O(log n)
		{
			index += N - 1;				
			sTree[index] = val;			
			while(index>1)				
			{
				index >>= 1;
				sTree[index] = Math.max(sTree[index<<1] ,sTree[index<<1|1]);		
			}
		}

		void update_range(int i, int j, int val) // O(log n)
		{
			update_range(1, 1, N, i, j, val);
		}

		void update_range(int node, int b, int e, int i, int j, int val) {
			if (i > e || j < b)
				return;
			if (b >= i && e <= j) {
				sTree[node] += (e - b + 1) * val;
				lazy[node] += val;
			} else {
				int mid = b + e >> 1;
				propagate(node, b, mid, e);
				update_range(node << 1, b, mid, i, j, val);
				update_range(node << 1 | 1, mid + 1, e, i, j, val);
				sTree[node] = sTree[node << 1] + sTree[node << 1 | 1];
			}

		}

		void propagate(int node, int b, int mid, int e) {
			lazy[node << 1] += lazy[node];
			lazy[node << 1 | 1] += lazy[node];
			sTree[node << 1] += (mid - b + 1) * lazy[node];
			sTree[node << 1 | 1] += (e - mid) * lazy[node];
			lazy[node] = 0;
		}

		int query(int i, int j) {
			return query(1, 1, N, i, j);
		}

		int query(int node, int b, int e, int i, int j) // O(log n)
		{
			if (i > e || j < b)
				return 0;
			if (b >= i && e <= j)
				return sTree[node];
			int mid = b + e >> 1;
			// propagate(node, b, mid, e);
			int q1 = query(node << 1, b, mid, i, j);
			int q2 = query(node << 1 | 1, mid + 1, e, i, j);
			return Math.max(q1,q2);

		}
	}

	static int memo[];


	static class UnionFind {
		int[] p, rank, setSize;
		int numSets;
		int max[];

		public UnionFind(int N) {
			p = new int[numSets = N];
			rank = new int[N];
			setSize = new int[N];
			for (int i = 0; i < N; i++) {
				p[i] = i;
				setSize[i] = 1;
			}
		}

		public int findSet(int i) {
			return p[i] == i ? i : (p[i] = findSet(p[i]));
		}

		public boolean isSameSet(int i, int j) {
			return findSet(i) == findSet(j);
		}

		public void unionSet(int i, int j) {
			if (isSameSet(i, j))
				return;
			numSets--;
			int x = findSet(i), y = findSet(j);
			if (rank[x] > rank[y]) {
				p[y] = x;
				setSize[x] += setSize[y];


			} else {
				p[x] = y;
				setSize[y] += setSize[x];
				if (rank[x] == rank[y])
					rank[y]++;


			}
		}

		

		public int numDisjointSets() {
			return numSets;
		}

		public int sizeOfSet(int i) {
			return setSize[findSet(i)];
		}
	}

	/**
	 * private static void trace(int i, int time) { if(i==n) return;
	 * 
	 * 
	 * long ans=dp(i,time);
	 * if(time+a[i].t<a[i].burn&&(ans==dp(i+1,time+a[i].t)+a[i].cost)) {
	 * 
	 * trace(i+1, time+a[i].t);
	 * 
	 * l1.add(a[i].idx); return; } trace(i+1,time);
	 * 
	 * }
	 **/

	static class incpair implements Comparable<incpair> {
		int a;
		long b;
		int idx;

		incpair(int a, long dirg, int i) {
			this.a = a;
			b = dirg;
			idx = i;
		}

		public int compareTo(incpair e) {
			return (int) (b - e.b);
		}
	}

	static class decpair implements Comparable<decpair> {
		int a;
		long b;
		int idx;

		decpair(int a, long dirg, int i) {
			this.a = a;
			b = dirg;
			idx = i;
		}

		public int compareTo(decpair e) {
			return (int) (e.b - b);
		}
	}

	static long allpowers[];

	static class Quad implements Comparable<Quad> {
		int u;
		int v;
		char state;
		int turns;

		public Quad(int i, int j, char c, int k) {
			u = i;
			v = j;
			state = c;
			turns = k;
		}

		public int compareTo(Quad e) {
			return (int) (turns - e.turns);
		}

	}

	static long dirg[][];
	static Edge[] driver;

	static int n;

	static class Edge implements Comparable<Edge> {
		int node;
		long cost;

		Edge(int a, long dirg) {
			node = a;
			cost = dirg;
		}

		public int compareTo(Edge e) {
			return (int) (cost - e.cost);
		}
	}

	static long manhatandistance(long x, long x2, long y, long y2) {
		return Math.abs(x - x2) + Math.abs(y - y2);
	}

	static long fib[];

	static long fib(int n) {
		if (n == 1 || n == 0) {
			return 1;
		}
		if (fib[n] != -1) {
			return fib[n];
		} else
			return fib[n] = ((fib(n - 2) % mod + fib(n - 1) % mod) % mod);
	}

	static class Pair implements Comparable<Pair> {
		int a, b;

		Pair(int x, int y) {
			a = x;
			b = y;
		}

		@Override
		public int compareTo(Pair c) {
			return b-c.b;
		}

	}

	static long[][] comb;

	static class Triple implements Comparable<Triple> {

		int l;
		int r;
		long cost;
		int idx;

		public Triple(int a, int b, long l1, int l2) {
			l = a;
			r = b;
			cost = l1;
			idx = l2;
		}

		public int compareTo(Triple x) {
			if (l != x.l || idx == x.idx)
				return l - x.l;
			return -idx;
		}

	}

	static TreeSet<Long> primeFactors(long N) // O(sqrt(N) / ln sqrt(N))
	{
		TreeSet<Long> factors = new TreeSet<Long>(); // take abs(N) in case of -ve integers
		int idx = 0, p = primes.get(idx);

		while (p * p <= N) {
			while (N % p == 0) {
				factors.add((long) p);
				N /= p;
			}
			if (primes.size() > idx + 1)
				p = primes.get(++idx);
			else
				break;
		}

		if (N != 1) // last prime factor may be > sqrt(N)
			factors.add(N); // for integers whose largest prime factor has a power of 1
		return factors;
	}

	static boolean visited[];

	/**
	 * static int bfs(int s) { Queue<Integer> q = new LinkedList<Integer>();
	 * q.add(s); int count=0; int maxcost=0; int dist[]=new int[n]; dist[s]=0;
	 * while(!q.isEmpty()) {
	 * 
	 * int u = q.remove(); if(dist[u]==k) { break; } for(Pair v: adj[u]) {
	 * maxcost=Math.max(maxcost, v.cost);
	 * 
	 * 
	 * 
	 * if(!visited[v.v]) {
	 * 
	 * visited[v.v]=true; q.add(v.v); dist[v.v]=dist[u]+1; maxcost=Math.max(maxcost,
	 * v.cost); } }
	 * 
	 * } return maxcost; }
	 **/
	public static boolean FindAllElements(int n, int k) {
		int sum = k;
		int[] A = new int[k];
		Arrays.fill(A, 0, k, 1);

		for (int i = k - 1; i >= 0; --i) {

			while (sum + A[i] <= n) {

				sum += A[i];
				A[i] *= 2;
			}
		}
		if (sum == n) {
			return true;
		} else
			return false;
	}

	static boolean vis2[][];

	static boolean f2 = false;

	static long[][] matMul(long[][] a2, long[][] b, int p, int q, int r) // C(p x r) = A(p x q) x (q x r) -- O(p x q x
																			// r)
	{
		long[][] C = new long[p][r];
		for (int i = 0; i < p; ++i) {
			for (int j = 0; j < r; ++j) {
				for (int k = 0; k < q; ++k) {
					C[i][j] = (C[i][j] + (a2[i][k] % mod * b[k][j] % mod)) % mod;
					C[i][j] %= mod;
				}

			}
		}
		return C;
	}

	static ArrayList<Pair> a1[];
	static int memo1[];

	static boolean vis[];
	static TreeSet<Integer> set = new TreeSet<Integer>();

	static long modPow(long ways, long count, long mod) // O(log e)
	{
		ways %= mod;
		long res = 1;
		while (count > 0) {
			if ((count & 1) == 1)
				res = (res * ways) % mod;
			ways = (ways * ways) % mod;
			count >>= 1;
		}
		return res % mod;
	}

	static long gcd(long ans, long b) {
		if (b == 0) {
			return ans;
		}
		return gcd(b, ans % b);
	}

	static int[] isComposite;
	static int[] valid;

	static ArrayList<Integer> primes;
	static ArrayList<Integer> l1;

	
	static TreeSet<Integer> primus = new TreeSet<Integer>();
	
	static void sieveLinear(int N)
	{
		int[] lp = new int[N + 1];								//lp[i] = least prime divisor of i
		for(int i = 2; i <= N; ++i)
		{
			if(lp[i] == 0)
			{
				primus.add(i);
				lp[i] = i;
			}
			int curLP = lp[i];
			for(int p: primus)
				if(p > curLP || p * i > N)
					break;
				else
					lp[p * i] = i;
		}
	}
	

	public static int[] schuffle(int[] p) {
		for (int i = 0; i < p.length; i++) {
			int x = (int) (Math.random() * p.length);
			int temp = p[x];
			p[x] = p[i];
			p[i] = temp;
		}
		return p;
	}

	static int V;
	static long INF = (long) 1E16;

	static class Edge2 {
		int node;
		long cost;
		long next;

		Edge2(int a, int c, Long long1) {
			node = a;
			cost = long1;
			next = c;
		}

	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}

		public int[] nxtArr(int n) throws IOException {
			int[] ans = new int[n];
			for (int i = 0; i < n; i++)
				ans[i] = nextInt();
			return ans;
		}
	}
}