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: /**
   * check out my youtube channel sh0rkyboy
   * https://tinyurl.com/zdxe2y4z
   * I do screencasts, solutions, and other fun stuff in the future
   */
  
  import java.util.*;
  import java.io.*;
  import static java.lang.Math.min;
  import static java.lang.Math.abs;
  import static java.lang.Math.max;
  public class EdE {
  	static long[] mods = {1000000007, 998244353, 1000000009};
  	static long mod = mods[0];
  	public static MyScanner sc;
      public static PrintWriter out;
  	public static void main(String[] largewang) throws Exception{
  		// TODO Auto-generated method stub
   		sc = new MyScanner();
   		out = new PrintWriter(System.out);
   		int t = sc.nextInt();
   		while (t-->0) {
   			int n = sc.nextInt();
   			int[] arr = readArrayInt(n);
   			int height = 1;
   			int index = n-1;
   			int start = arr[index];
   			int max = -1;
   			for(int j : arr){
   				max = max(j, max);
   			}
   			int count = 0;
   			while(start != max){
   				count++;
  	 			while(arr[index] <= start) {
  	 				index --;
  	 			}
  	 			start = arr[index];
   			}
   			out.println(count);
   		}
   		out.close();
   		
   	}
  	
  	public static void sort(int[] array){
  		ArrayList<Integer> copy = new ArrayList<>();
  		for (int i : array)
  			copy.add(i);
  		Collections.sort(copy);
  		for(int i = 0;i<array.length;i++)
  			array[i] = copy.get(i);
  	}
  	static String[] readArrayString(int n){
  		String[] array = new String[n];
  		for(int j =0 ;j<n;j++)
  			array[j] = sc.next();
  		return array;
  	}
  	static int[] readArrayInt(int n){
      	int[] array = new int[n];
      	for(int j = 0;j<n;j++)
      		array[j] = sc.nextInt();
      	return array;
      }
  	static int[] readArrayInt1(int n){
  		int[] array = new int[n+1];
  		for(int j = 1;j<=n;j++){
  			array[j] = sc.nextInt();
  		}
  		return array;
  	}
  	static long[] readArrayLong(int n){
  		long[] array = new long[n];
  		for(int j =0 ;j<n;j++)
  			array[j] = sc.nextLong();
  		return array;
  	}
  	static double[] readArrayDouble(int n){
  		double[] array = new double[n];
  		for(int j =0 ;j<n;j++)
  			array[j] = sc.nextDouble();
  		return array;
  	}
  	static int minIndex(int[] array){
  		int minValue = Integer.MAX_VALUE;
  		int minIndex = -1;
  		for(int j = 0;j<array.length;j++){
  			if (array[j] < minValue){
  				minValue = array[j];
  				minIndex = j;
  			}
  		}
  		return minIndex;
  	}
  	static int minIndex(long[] array){
  		long minValue = Long.MAX_VALUE;
  		int minIndex = -1;
  		for(int j = 0;j<array.length;j++){
  			if (array[j] < minValue){
  				minValue = array[j];
  				minIndex = j;
  			}
  		}
  		return minIndex;
  	}
  	static int minIndex(double[] array){
  		double minValue = Double.MAX_VALUE;
  		int minIndex = -1;
  		for(int j = 0;j<array.length;j++){
  			if (array[j] < minValue){
  				minValue = array[j];
  				minIndex = j;
  			}
  		}
  		return minIndex;
  	}
  	public static int power(long x, long y, long mod){
  		long ans = 1;
  		while(y>0){
  			if (y%2==1)
  				ans = (ans*x)%mod;
  			x = (x*x)%mod;
  			y/=2;
  		}
  		return (int)(ans);
  	}
  	static void verdict(boolean a){
          out.println(a ? "YES" : "NO");
      }
      public static class MyScanner {
          BufferedReader br;
          StringTokenizer st;
          public MyScanner() {
              br = new BufferedReader(new InputStreamReader(System.in));
          }
          String next() { 
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } 
                  catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
          int nextInt() {
              return Integer.parseInt(next());
          }
          long nextLong() {
              return Long.parseLong(next());
          }
          double nextDouble() {
              return Double.parseDouble(next());
          }
          String nextLine() {
              String str = "";
              try{
                  str = br.readLine();
              } 
              catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
          
      }	
  }
  
  //StringJoiner sj = new StringJoiner(" "); 
  //sj.add(strings)
  //sj.toString() gives string of those stuff w spaces or whatever that sequence is