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.util.stream.Collectors;
  import java.io.*;
  import java.math.*;
  
  
  
  public class A3 {
  	public static FastScanner sc;
  	public static PrintWriter pw;
  	public static StringBuilder sb;
  	
      public static int MOD= 1000000007;
  	public static class FastScanner {
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st=new StringTokenizer("");
  		String next() {
  			while (!st.hasMoreTokens())
  				try { 
  					st=new StringTokenizer(br.readLine());				               
  	             } catch (IOException e) {}
  				return st.nextToken();
  			}
  		
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		long nextLong() {
  			return Long.parseLong(next());
  		}
      }
  	
  	public static void printList(List<Long> al) {
     	 System.out.println(al.stream().map(it -> it.toString()).collect(Collectors.joining(" ")));
  	}
  	public static void printList(Deque<Long> al) {
    	 System.out.println(al.stream().map(it -> it.toString()).collect(Collectors.joining(" ")));
  	}
  	public static void printArr(int[] arr) {
     	System.out.println(Arrays.toString(arr).trim().replace("[", "").replace("]","").replace(","," "));
  	}
      
      public static long gcd(long a, long b) {
  		if(b==0) return a;
  		return gcd(b,a%b);
  	}
  	
  	public static long lcm(long a, long b) {
  		return((a*b)/gcd(a,b));
  	}
  	
  	public static void decreasingOrder(ArrayList<Long> al) {
  		Comparator<Long> c = Collections.reverseOrder();
  	    Collections.sort(al,c);
  	}
      
      public static boolean[] sieveOfEratosthenes(int n) {
  		boolean isPrime[] = new boolean[n+1];
  		Arrays.fill(isPrime, true);
  		isPrime[0]=false;
  		isPrime[1]=false;	
  		for(int i=2;i*i<n;i++) {
  			for(int j=2*i;j<n;j+=i) {
  				isPrime[j]=false;
  			}
  		}
  		return isPrime;
  	}
      
      public static  long nCr(long N, long R) {
          double result=1;
      	for(int i=1;i<=R;i++) result=((result*(N-R+i))/i);
      	return (long) result;
      }
      
      public static long fastPow(long a, long b, int n) {
  		long result=1;
  		
  		while(b>0) {
  			if((b&1)==1)
  				result=(result*a %n)%n;
  			
  			a=(a%n * a%n)%n;
  			b>>=1;
  		}
  		return result;
      }
      
      public static int BinarySearch(long[] arr, long key) {
  		int low=0;
  		int high=arr.length-1;
  		while(low<=high) {
  			int mid=(low+high)/2;
  			if(arr[mid]==key)
  				return mid;
  			else if(arr[mid]>key)
  				high=mid-1;
  			else
  				low=mid+1;
  		}
  		return low; //High=low-1
  	}
          
      public static void solve(int t) {
      	int n=sc.nextInt();
      	int[] arr = new int[n];
      	int max=Integer.MIN_VALUE;
      	int index=-1;
      	for(int i=0;i<n;i++) {
      		arr[i]=sc.nextInt();
      		if(arr[i]>=max) {
      			max=arr[i];
      			index=i;
      		}
      	}
      	int count=0;
      	int temp=Integer.MIN_VALUE;
      	
      	for(int i=n-1;i>=index+1;i--) {
      		if(arr[i]!=max) {
      			if(arr[i]>temp) {
      				temp=arr[i];
      				count++;
      			}
      		}
      	}
      	
      	
      	
      	System.out.println(count);
      	
      	
      }
      
  	public static void main(String[] args) {
  		sc = new FastScanner();
  		pw = new PrintWriter(new OutputStreamWriter(System.out));
  		sb= new StringBuilder("");
  		int t=sc.nextInt();
  		for(int i=1;i<=t;i++) solve(i);
  		System.out.println(sb);
  	}
  }
  
  
  
  