description: You are given an array a of length n.
  
  Let's define the eversion operation. Let x = a_n. Then array a is partitioned into two parts: left and right. The left part contains the elements of a that are not greater than x (≤ x). The right part contains the elements of a that are strictly greater than x (> x). The order of elements in each part is kept the same as before the operation, i. e. the partition is stable. Then the array is replaced with the concatenation of the left and the right parts.
  
  For example, if the array a is [2, 4, 1, 5, 3], the eversion goes like this: [2, 4, 1, 5, 3] → [2, 1, 3], [4, 5] → [2, 1, 3, 4, 5].
  
  We start with the array a and perform eversions on this array. We can prove that after several eversions the array a stops changing. Output the minimum number k such that the array stops changing after k eversions.
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
  
  The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5).
  
  The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
  
  It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
  
  Output
  
  For each test case print a single integer k — the number of eversions after which the array stops changing.
  
  Example
  
  Input
  
  
  3
  5
  2 4 1 5 3
  5
  5 3 2 4 1
  4
  1 1 1 1
  
  
  Output
  
  
  1
  2
  0
  
  Note
  
  Consider the fist example.
  
    * The first eversion: a = [1, 4, 2, 5, 3], x = 3. [2, 4, 1, 5, 3] → [2, 1, 3], [4, 5] → [2, 1, 3, 4, 5]. 
    * The second and following eversions: a = [2, 1, 3, 4, 5], x = 5. [2, 1, 3, 4, 5] → [2, 1, 3, 4, 5], [] → [2, 1, 3, 4, 5]. This eversion does not change the array, so the answer is 1. 
  
  
  
  Consider the second example. 
  
    * The first eversion: a = [5, 3, 2, 4, 1], x = 1. [5, 3, 2, 4, 1] → [1], [5, 3, 2, 4] → [1, 5, 3, 2, 4]. 
    * The second eversion: a = [1, 5, 3, 2, 4], x = 4. [1, 5, 3, 2, 4] → [1, 3, 2, 4], [5] → [1, 3, 2, 4, 5]. 
    * The third and following eversions: a = [1, 3, 2, 4, 5], x = 5. [1, 3, 2, 4, 5] → [1, 3, 2, 4, 5], [] → [1, 3, 2, 4, 5]. This eversion does not change the array, so the answer is 2. 
solution: import java.util.*;
  import java.lang.*;
  import java.io.*;
  
  public class Main
  {
  	public static void main (String[] args) throws java.lang.Exception
  	{
  		Reader scan = new Reader();
  		int tc = scan.nextInt();
  		for (int i = 0; i < tc; i++){
  		    int n = scan.nextInt();
  		    ArrayList <Long> arr = new ArrayList <Long>();
  		    for (int j = 0; j < n; j++){
  		        arr.add(scan.nextLong());
  		    }
              System.out.println(solve(arr));
  		}
  	}
  
      public static long solve (ArrayList <Long> arr){
          //long max = max(arr);
          long last = arr.get(arr.size()-1);
          long result = 0;
          for (int i = arr.size()-1; i >= 0; i--){
              if (arr.get(i) > last){
                  last = arr.get(i);
                  result++;
              }
          }
          return result;
      }
      
      public static long max (ArrayList <Long> arr){
          long max = 0;
          for (int i = 0; i < arr.size(); i++){
              max = Math.max(arr.get(i), max);
          }
          return max;
      }
  
      public static long modPow(long x, long n){
          long mod = 1000000007;
          long result = 1;
          while (n > 0){
              if (n % 2 != 0)
                  result = (result % mod) * (x % mod) % mod;
              x *= x % mod;
              n /= 2;
          }
          return result;
      }
      
  	
  	public static void printarr(int[]arr){
  	    for (int i = 0; i < arr.length; i++){
  	        System.out.print(arr[i] + " ");
  	    }
  	    System.out.println();
  	}
  
      public static void printarr(ArrayList <Integer> arr){
  	    for (int i = 0; i < arr.size(); i++){
  	        System.out.print(arr.get(i) + " ");
  	    }
  	    System.out.println();
  	}
  	
  	static class Reader {
          final private int BUFFER_SIZE = 1 << 64;
          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[1000000]; // line length
              int cnt = 0, c;
              while ((c = read()) != -1) {
                  if (c == '\n') {
                      if (cnt != 0) {
                          break;
                      }
                      else {
                          continue;
                      }
                  }
                  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();
          }
      }
  }
  