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.util.*;
import java.lang.*;
import java.io.*;
import java.math.BigInteger; 

public class Codechef
{
    static long fac[] = new long[1000001];
    static void perm()
    {
        fac[0]=1;
        for (int x=1;x<=1000000;x++)
        fac[x]=fac[x-1]*x;
    }
    static long power(long x, long y, long p) 
    { 
          
        // Initialize result 
        long res = 1; 
      
        // Update x if it is more than or 
        // equal to p 
        x = x % p; 
                      
        while (y > 0) 
        { 
              
            // If y is odd, multiply x 
            // with result 
            if (y % 2 == 1) 
                res = (res * x) % p; 
      
            // y must be even now 
            y = y >> 1; // y = y/2 
            x = (x * x) % p; 
        } 
          
        return res; 
    } 
    // Returns n^(-1) mod p 
    static long modInverse(long n, long p) 
    { 
        return power(n, p-2, p); 
    } 
      
    // Returns nCr % p using Fermat's 
    // little theorem. 
    static long nPr(int n, int r, 
                                    long p) 
    { 
          
        // Base case 
        if (r == 0) 
            return 1; 
      
        // Fill factorial array so that we 
        // can find all factorial of r, n 
        // and n-r 
      
        return ((fac[n] % p) * modInverse(fac[n-r], p) 
                                    % p) % p; 
    }
static long mod(String num, long a) 
    { 
          
        // Initialize result 
        long res = 0; 
  
        // One by one process all digits of 'num' 
        for (int i = 0; i < num.length(); i++) 
            res = (res * 10 + (int)num.charAt(i) 
                                     - '0') % a; 
  
        return res; 
    }
    static boolean check (int a)
    {
        String s=Integer.toString(a);
        int f[] = new int[10];
        for (int x=0;x<s.length();x++)
        {
            f[s.charAt(x)-'0']++;
            if (f[s.charAt(x)-'0']>1)
            return (false);
        }
        return (true);
    }
    static long lcm(int[] element_array) 
    { 
        long lcm_of_array_elements = 1; 
        int divisor = 2; 
          
        while (true) { 
            int counter = 0; 
            boolean divisible = false; 
              
            for (int i = 0; i < element_array.length; i++) { 
  
                // lcm_of_array_elements (n1, n2, ... 0) = 0. 
                // For negative number we convert into 
                // positive and calculate lcm_of_array_elements. 
  
                if (element_array[i] == 0) { 
                    return 0; 
                } 
                else if (element_array[i] < 0) { 
                    element_array[i] = element_array[i] * (-1); 
                } 
                if (element_array[i] == 1) { 
                    
                    
                    counter++; 
                } 
  
                // Divide element_array by devisor if complete 
                // division i.e. without remainder then replace 
                // number with quotient; used for find next factor 
                if (element_array[i] % divisor == 0) { 
                    divisible = true; 
                    element_array[i] = element_array[i] / divisor; 
                } 
            } 
  
            // If divisor able to completely divide any number 
            // from array multiply with lcm_of_array_elements 
            // and store into lcm_of_array_elements and continue 
            // to same divisor for next factor finding. 
            // else increment divisor 
            if (divisible) { 
                lcm_of_array_elements = lcm_of_array_elements * divisor; 
            } 
            else { 
                divisor++; 
            } 
  
            // Check if all element_array is 1 indicate  
            // we found all factors and terminate while loop. 
            if (counter == element_array.length) { 
                return lcm_of_array_elements; 
            } 
        } 
    }
    static boolean find(int N)
    {
        String r="";
        int rem=1%N;
        HashMap<Integer, Integer> h = new HashMap<Integer, Integer>();
        while ((rem!=0) && !h.containsKey(rem))
        {
            h.put(rem,r.length());
            rem=rem*10;
            r=r+(rem/N);
            rem=rem%N;
        }
        if (rem==0)
        return (false);
        else
        return (true);
    }
    static String sorts(String inputString) 
    { 
        // convert input string to char array 
        char tempArray[] = inputString.toCharArray(); 
          
        // sort tempArray 
        Arrays.sort(tempArray); 
          
        // return new sorted string 
        return new String(tempArray); 
    }
    static long qrt(double d)
    {
        long k=(long)(d);
        double x=-1+Math.sqrt(1+(8*d));
        x=x/2;
        long w=(long)(Math.ceil(x));
        long sum=w*(w+1);
        sum=sum/2;
        //System.out.println (w+"GDSGDS");
        while (sum%2!=k%2)
        {
            w++;
            sum=w*(w+1);
            sum=sum/2;
        }
        return (w);
    }
    static int knapSack(long W, int wt[], int val[], int n) 
     { 
        // Base Case 
    if (n <=0 || W == 0) 
        return 0; 
    if (wt[n-1] > W) 
       return knapSack(W, wt, val, n-1); 
    else 
    return Math.max( val[n-1] + knapSack(W-wt[n-1], wt, val, n-2), knapSack(W, wt, val, n-1)); 
      } 
    static boolean index(int x, int y, int N, int M)
    {
        if (x>=0 && y>=0 && x<N && y<M)
        return (true);
        else
        return (false);
    }
    static boolean check (String a)
    {
        int len=a.length();
        char z;
        int l=0;
        int r=len-1;int p=0;
        while (l<=r && p==0)
        {
            if (a.charAt(l)==a.charAt(r))
            {l++;r--;}
            else 
            p=1;
        }
        if (p==0)
        return (true);
        else
        return (false);
    }
	public static void main (String[] args) throws java.lang.Exception
	{
	    Reader in = new Reader();
	    int T=in.nextInt();
	    //String k=in.readLine();
	    for (int Q=1;Q<=T;Q++)
	    {
	        int N=in.nextInt();
	        //HashMap<Integer, Integer> h = new HashMap<Integer, Integer>();
	        int king[] = new int[N+1];
	        int kin;
	        int p=0;
	        int w=0;
	        int ki=-1;int y=0;
	        for (int n=1;n<=N;n++)
	        {
	            int K=in.nextInt();
	            p=0;
	            for (int k=1;k<=K;k++)
	            {
	                kin = in.nextInt();
	                if (king[kin]==0 && p==0)
	                {king[kin]=1;p=1;}
	            }
	            if (p==0 && w==0)
	            {w=1;ki=n;}
	        }
	        if (w==0)
	        System.out.println ("OPTIMAL");
	        else
	        {
	            for (int x=1;x<=N;x++)
	            if (king[x]==0)
	            {System.out.println ("IMPROVE\n"+ki+" "+x);break;}
	        }
	     //   System.out.println (w+"ddfd "+Q);
	    }
	}
  static class Reader
    {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;
 
        public Reader()
        {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }
 
        public Reader(String file_name) throws IOException
        {
            din = new DataInputStream(new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }
 
        public String readLine() throws IOException
        {
            byte[] buf = new byte[64]; // line length
            int cnt = 0, c;
            while ((c = read()) != -1)
            {
                if (c == '\n')
                    break;
                buf[cnt++] = (byte) c;
            }
            return new String(buf, 0, cnt);
        }
 
        public int nextInt() throws IOException
        {
            int ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do
            {
                ret = ret * 10 + c - '0';
            }  while ((c = read()) >= '0' && c <= '9');
 
            if (neg)
                return -ret;
            return ret;
        }
 
        public long nextLong() throws IOException
        {
            long ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }
 
        public double nextDouble() throws IOException
        {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
 
            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');
 
            if (c == '.')
            {
                while ((c = read()) >= '0' && c <= '9')
                {
                    ret += (c - '0') / (div *= 10);
                }
            }
 
            if (neg)
                return -ret;
            return ret;
        }
 
        private void fillBuffer() throws IOException
        {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }
 
        private byte read() throws IOException
        {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }
 
        public void close() throws IOException
        {
            if (din == null)
                return;
            din.close();
        }
    }
}