Monocarp had a tree which consisted of n vertices and was rooted at vertex 1. He decided to study BFS ([Breadth-first search](https://en.wikipedia.org/wiki/Breadth-first_search)), so he ran BFS on his tree, starting from the root. BFS can be described by the following pseudocode:
    
    
    a = [] # the order in which vertices were processed  
    q = Queue()  
    q.put(1) # place the root at the end of the queue  
    while not q.empty():  
        k = q.pop() # retrieve the first vertex from the queue  
        a.append(k) # append k to the end of the sequence in which vertices were visited  
        for y in g[k]: # g[k] is the list of all children of vertex k, sorted in ascending order  
            q.put(y)  
    

Monocarp was fascinated by BFS so much that, in the end, he lost his tree. Fortunately, he still has a sequence of vertices, in which order vertices were visited by the BFS algorithm (the array a from the pseudocode). Monocarp knows that each vertex was visited exactly once (since they were put and taken from the queue exactly once). Also, he knows that all children of each vertex were viewed in ascending order.

Monocarp knows that there are many trees (in the general case) with the same visiting order a, so he doesn't hope to restore his tree. Monocarp is okay with any tree that has minimum height.

The height of a tree is the maximum depth of the tree's vertices, and the depth of a vertex is the number of edges in the path from the root to it. For example, the depth of vertex 1 is 0, since it's the root, and the depth of all root's children are 1.

Help Monocarp to find any tree with given visiting order a and minimum height.

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

The first line of each test case contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of vertices in the tree.

The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n; a_i ≠ a_j; a_1 = 1) — the order in which the vertices were visited by the BFS algorithm.

It's guaranteed that the total sum of n over test cases doesn't exceed 2 ⋅ 10^5.

Output

For each test case print the minimum possible height of a tree with the given visiting order a.

Example

Input


3
4
1 4 3 2
2
1 2
3
1 2 3


Output


3
1
1

Note

In the first test case, there is only one tree with the given visiting order: 

<image>

In the second test case, there is only one tree with the given visiting order as well: 

<image>

In the third test case, an optimal tree with the given visiting order is shown below: 

<image>
// Don't place your source in a package
import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.*; 
 
 /*
 
 6666666666666			 6666666666666                     
 6						 6  											   			
 6						 6                                      
 6						 6                                      
 6						 6                                       
 6                       6                                                 
 6666666666666           6666666666666                     
 6		     6           6		     6
 6			 6	         6			 6 
 6			 6           6			 6
 6		     6	         6		     6	
 6666666666666           6666666666666
 
 
 */
 
 
 
 
 
// Please name your class Main
public class Main {
	
	//static StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
 
    /*static int read() throws IOException {
        in.nextToken();
        return (int) in.nval;
    }
    static String readString() throws IOException {
        in.nextToken();
        return in.sval;
	}*/
	
	
	static Scanner in = new Scanner(System.in);
	
	public static void main (String[] args) throws java.lang.Exception {
	    
		//InputReader in = new InputReader(System.in);
		PrintWriter out = new PrintWriter(System.out);
		
 
		int T=Int();
		
		for(int t=0;t<T;t++){
			int n=Int();
			int A[]=new int[n];
			for(int i=0;i<n;i++){
				A[i]=Int();
			}
			Solution sol=new Solution();
			sol.solution(A);
		}
 
		out.flush();
	}
	
	
	
	public static long Long(){
		return in.nextLong();
	}
	public static int Int(){
		return in.nextInt();
	}
	public static String Str(){
		return in.next();
	}
}
 
class Solution{
	//constant variable
	final int MAX=Integer.MAX_VALUE;
	final int MIN=Integer.MIN_VALUE;
	//Set<Integer>adjecent[];
	//////////////////////////////
 
	public void fastprint(PrintWriter out,String s){
		out.print(s);
	}
	
	public void fastprintln(PrintWriter out,String s){
		out.println(s);
	}


	
	public void solution(int A[]){
		int res=0;
		Queue<Integer>q=new LinkedList<>();
		q.add(0);
		
		int i=1;
		
		while(q.size()!=0){
			int level=q.poll();
			res=Math.max(res,level);
			int cnt=1;
			int j=i+1;
			while(j<A.length&&A[j]>A[j-1]){
				j++;
				cnt++;
			}
			for(int k=0;k<cnt;k++){
				q.add(level+1);
			}
			i=j;
			if(i>=A.length)break;
		}
		
		
		while(q.size()!=0){
			res=Math.max(res,q.poll());
		}
		msg(res+"");
	}
	
	
	



 
 
 
 
	public void sort(int A[]){
		Arrays.sort(A);
	}
	
	public void swap(int A[],int l,int r){
		int t=A[l];
		A[l]=A[r];
		A[r]=t;
	}
	
 
	
	
	public long C(long fact[],int i,int j){ // C(20,3)=20!/(17!*3!)
		// take   a/b     where a=20!  b=17!*3!
		if(j>i)return 0;
		if(j==0)return 1;
		long mod=1000000007;
		long a=fact[i];
		long b=((fact[i-j]%mod)*(fact[j]%mod))%mod;
		BigInteger B= BigInteger.valueOf(b);
		long binverse=B.modInverse(BigInteger.valueOf(mod)).longValue();
		return ((a)*(binverse%mod))%mod;
	}
	
	public long gcd(long a, long b) 
    { 
        if (a == 0) 
            return b; 
          
        return gcd(b%a, a); 
    } 
	
	 
	
	//map operation
	public void put(Map<Integer,Integer>map,int i){
		if(!map.containsKey(i))map.put(i,0);
		map.put(i,map.get(i)+1);
	}
	
	public void delete(Map<Integer,Integer>map,int i){
		map.put(i,map.get(i)-1);
		if(map.get(i)==0)map.remove(i);
	}
 
 
 
 
 
	
 
 
	/*public void tarjan(int p,int r){
		if(cut)return;
		List<Integer>childs=adjecent[r];
		dis[r]=low[r]=time;
		time++;
		
		//core for tarjan
		int son=0;
		for(int c:childs){
			if(ban==c||c==p)continue;
			if(dis[c]==-1){
				son++;
				tarjan(r,c);
				low[r]=Math.min(low[r],low[c]);
				if((r==root&&son>1)||(low[c]>=dis[r]&&r!=root)){
					cut=true;
					return;
				}
			}else{
				if(c!=p){
					low[r]=Math.min(low[r],dis[c]);
				}
			}
		}
		
	}*/
	
	
 
	
	//helper function I would use
	public void remove(Map<Integer,Integer>map,int i){
		map.put(i,map.get(i)-1);
		if(map.get(i)==0)map.remove(i);
	}
	
	
	
	public void ascii(String s){
		for(char c:s.toCharArray()){
			System.out.print((c-'a')+"  ");
		}
		msg("");
	}
	
	public int flip(int i){
		if(i==0)return 1;
		else return 0;
	}
	
	public boolean[] primes(int n){
		boolean A[]=new boolean[n+1];
		for(int i=2;i<=n;i++){
			if(A[i]==false){
				for(int j=i+i;j<=n;j+=i){
					A[j]=true;
				}
			}
		}
		return A;
	}
	
	public void msg(String s){
		System.out.println(s);
	}
	
	public void msg1(String s){
		System.out.print(s);
	}
	
	public int[] kmpPre(String p){
		int pre[]=new int[p.length()];
		int l=0,r=1;
		while(r<p.length()){
			if(p.charAt(l)==p.charAt(r)){
				pre[r]=l+1;
				l++;r++;
			}else{
				if(l==0)r++;
				else l=pre[l-1];
			}
		}
		return pre;
	}
	
	public boolean isP(String s){
		int l=0,r=s.length()-1;
		while(l<r){
			if(s.charAt(l)!=s.charAt(r))return false;
			l++;r--;
		}
		return true;
	}
	
	public int find(int nums[],int x){//union find => find method
		if(nums[x]==x)return x;
		int root=find(nums,nums[x]);
		nums[x]=root;
		return root;
	}
	
 
	
	public int get(int A[],int i){
		if(i<0||i>=A.length)return 0;
		return A[i];
	}
	public int[] copy1(int A[]){
		int a[]=new int[A.length];
		for(int i=0;i<a.length;i++)a[i]=A[i];
		return a;
	}
	
	
	public void print1(long A[]){
		for(long i:A)System.out.print(i+"  ");
		System.out.println();
	}
	public void print2(int A[][]){
		for(int i=0;i<A.length;i++){
			for(int j=0;j<A[0].length;j++){
				System.out.print(A[i][j]+"   ");
			}System.out.println();
		}
	}
	
	public int min(int a,int b){
		return Math.min(a,b);
	}
	
	
	public int[][] matrixdp(int[][] grid) {
        if(grid.length==0)return new int[][]{};
        int res[][]=new int[grid.length][grid[0].length];
        for(int i=0;i<grid.length;i++){
            for(int j=0;j<grid[0].length;j++){
                res[i][j]=grid[i][j]+get(res,i-1,j)+get(res,i,j-1)-get(res,i-1,j-1);
            }
        }
        return res;
    }
    
    public int get(int grid[][],int i,int j){
        if(i<0||j<0||i>=grid.length||j>=grid[0].length)return 0;
        return grid[i][j];
    }
	
	public int[] suffixArray(String s){
		int n=s.length();
		Suffix A[]=new Suffix[n];
		
		for(int i=0;i<n;i++){
			A[i]=new Suffix(i,s.charAt(i)-'a',0);
		}
		
		for(int i=0;i<n;i++){
			if(i==n-1){
				A[i].next=-1;
			}else{
				A[i].next=A[i+1].rank;
			}
		}
		Arrays.sort(A);
		
		for(int len=4;len<A.length*2;len<<=1){
			int in[]=new int[A.length];
			int rank=0;
			int pre=A[0].rank;
			A[0].rank=rank;
			in[A[0].index]=0;
			for(int i=1;i<A.length;i++){//rank for the first two letter
				if(A[i].rank==pre&&A[i].next==A[i-1].next){
					pre=A[i].rank;
					A[i].rank=rank;
				}else{
					pre=A[i].rank;
					A[i].rank=++rank;
				}
				in[A[i].index]=i;
			}
			
			for(int i=0;i<A.length;i++){
				int next=A[i].index+len/2;
				if(next>=A.length){
					A[i].next=-1;
				}else{
					A[i].next=A[in[next]].rank;
				}
			}
			
			
			Arrays.sort(A);
		}
		
		
		int su[]=new int[A.length];
		for(int i=0;i<su.length;i++){
			su[i]=A[i].index;
		}
		return su;
	}
	
}
 
//suffix array Struct
class Suffix implements Comparable<Suffix>{
	int index;
	int rank;
	int next;
	public Suffix(int i,int rank,int next){
		this.index=i;
		this.rank=rank;
		this.next=next;
	}
	
	@Override
	public int compareTo(Suffix other) {
		if(this.rank==other.rank){
			return this.next-other.next;
		}
		return this.rank-other.rank;
	}
	
	public String toString(){
		return this.index+"   "+this.rank+"   "+this.next+"  ";
	}
}
 
 
 
class Wrapper implements Comparable<Wrapper>{
	int spf;int cnt;
	public Wrapper(int spf,int cnt){
		this.spf=spf;
		this.cnt=cnt;
	}
		
	@Override
	public int compareTo(Wrapper other) {
		return this.spf-other.spf;
	}
}
 
 
 
    class Node{//what the range would be for that particular node
		boolean state=false;
        int l=0,r=0;
		int ll=0,rr=0;
        public Node(boolean state){
            this.state=state;
        }
    }
	
	
	
	
	class Seg1{
			int A[];
			public Seg1(int A[]){
				this.A=A;
			}
			
			public void update(int left,int right,int val,int s,int e,int id){
				if(left<0||right<0||left>right)return;
				if(left==s&&right==e){
					
					A[id]+=val;
					return;
				}
				int mid=s+(e-s)/2; //[s,mid] [mid+1,e]
				
				if(left>=mid+1){
					update(left,right,val,mid+1,e,id*2+2);
				}else if(right<=mid){
					update(left,right,val,s,mid,id*2+1);
				}else{
					update(left,mid,val,s,mid,id*2+1);
					update(mid+1,right,val,mid+1,e,id*2+2);
				}
			}
			
			public int query(int i,int add,int s,int e,int id){
				
				if(s==e&&i==s){
					return A[id]+add;
				}
				
				int mid=s+(e-s)/2; //[s,mid] [mid+1,e]
				
				if(i>=mid+1){
					return query(i,A[id]+add,mid+1,e,id*2+2);
				}else{
					return query(i,A[id]+add,s,mid,id*2+1);
				}
			}
	}
 
 
	 class MaxFlow{
		
 
		  public static List<Edge>[] createGraph(int nodes) {
			List<Edge>[] graph = new List[nodes];
			for (int i = 0; i < nodes; i++)
			  graph[i] = new ArrayList<>();
			return graph;
		  }
 
		  public static void addEdge(List<Edge>[] graph, int s, int t, int cap) {
			graph[s].add(new Edge(t, graph[t].size(), cap));
			graph[t].add(new Edge(s, graph[s].size() - 1, 0));
		  }
 
		  static boolean dinicBfs(List<Edge>[] graph, int src, int dest, int[] dist) {
			Arrays.fill(dist, -1);
			dist[src] = 0;
			int[] Q = new int[graph.length];
			int sizeQ = 0;
			Q[sizeQ++] = src;
			for (int i = 0; i < sizeQ; i++) {
			  int u = Q[i];
			  for (Edge e : graph[u]) {
				if (dist[e.t] < 0 && e.f < e.cap) {
				  dist[e.t] = dist[u] + 1;
				  Q[sizeQ++] = e.t;
				}
			  }
			}
			return dist[dest] >= 0;
		  }
 
		  static int dinicDfs(List<Edge>[] graph, int[] ptr, int[] dist, int dest, int u, int f) {
			if (u == dest)
			  return f;
			for (; ptr[u] < graph[u].size(); ++ptr[u]) {
			  Edge e = graph[u].get(ptr[u]);
			  if (dist[e.t] == dist[u] + 1 && e.f < e.cap) {
				int df = dinicDfs(graph, ptr, dist, dest, e.t, Math.min(f, e.cap - e.f));
				if (df > 0) {
				  e.f += df;
				  graph[e.t].get(e.rev).f -= df;
				  return df;
				}
			  }
			}
			return 0;
		  }
 
		  public static int maxFlow(List<Edge>[] graph, int src, int dest) {
			int flow = 0;
			int[] dist = new int[graph.length];
			while (dinicBfs(graph, src, dest, dist)) {
			  int[] ptr = new int[graph.length];
			  while (true) {
				int df = dinicDfs(graph, ptr, dist, dest, src, Integer.MAX_VALUE);
				if (df == 0)
				  break;
				flow += df;
			  }
			}
			return flow;
		  }
		 
	 }
	 
	 	 class Edge {
			int t, rev, cap, f;
 
			public Edge(int t, int rev, int cap) {
			  this.t = t;
			  this.rev = rev;
			  this.cap = cap;
			}
		  }
		  
	   class Seg{
        int l,r;
        int min=Integer.MAX_VALUE;
        Seg left=null,right=null;
		int tree[];
        public Seg(int l,int r){
            this.l=l;
            this.r=r;
            if(l!=r){
                int mid=l+(r-l)/2;
                if(l<=mid)left=new Seg(l,mid);
                if(r>=mid+1)right=new Seg(mid+1,r);
                if(left!=null)min=Math.min(left.min,min);
                if(right!=null)min=Math.min(right.min,min);
            }else{
                min=tree[l];
            }
        }
        public int query(int s,int e){
            if(l==s&&r==e){
                return min;
            }
            int mid=l+(r-l)/2; //left :  to mid-1,
            if(e<=mid){
                return left.query(s,e);
            }
            else if(s>=mid+1){
                return right.query(s,e);
            }else{
                return Math.min(left.query(s,mid),right.query(mid+1,e));
            }
        }

        public void update(int index){
            if(l==r){
                min=tree[l];
                return;
            }
            int mid=l+(r-l)/2;
            if(index<=mid){
                left.update(index);
            }else{
                right.update(index);
            }
            this.min=Math.min(left.min,right.min);
        }
    }
	
    
		  
		  