Problem: You are given an array a consisting of n integers.

Let min(l, r) be the minimum value among a_l, a_{l + 1}, …, a_r and max(l, r) be the maximum value among a_l, a_{l + 1}, …, a_r.

Your task is to choose three positive (greater than 0) integers x, y and z such that:

  * x + y + z = n; 
  * max(1, x) = min(x + 1, x + y) = max(x + y + 1, n). 



In other words, you have to split the array a into three consecutive non-empty parts that cover the whole array and the maximum in the first part equals the minimum in the second part and equals the maximum in the third part (or determine it is impossible to find such a partition).

Among all such triples (partitions), you can choose any.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (3 ≤ n ≤ 2 ⋅ 10^5) — the length of a.

The second line of the test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.

It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).

Output

For each test case, print the answer: NO in the only line if there is no such partition of a that satisfies the conditions from the problem statement. Otherwise, print YES in the first line and three integers x, y and z (x + y + z = n) in the second line.

If there are several answers, you can print any.

Example

Input


6
11
1 2 3 3 3 4 4 3 4 2 1
8
2 9 1 7 3 9 4 1
9
2 1 4 2 4 3 3 1 2
7
4 2 1 1 4 1 4
5
1 1 1 1 1
7
4 3 4 3 3 3 4


Output


YES
6 1 4
NO
YES
2 5 2
YES
4 1 2
YES
1 1 3
YES
2 1 4

Solution (in CPP):

Solution:
#include<bits/stdc++.h>

using namespace std;

/*

        Bismillahir Rahmanir Rahim
        Problem :
        Problem Link :
        Topics :
        Solver : Masud Parves
        I Love Code More than Sharks Love Blood <3
*/

#define ff first
#define ss second
#define mp make_pair
#define SZ(a)           (int)a.size()
#define all(a)  a.begin(), a.end()
#define allr(a)  a.rbegin(), a.rend()


#define sf(a)           scanf("%d",&a)
#define sff(a,b)        scanf("%d %d",&a,&b)
#define sfff(a,b,c)     scanf("%d %d %d",&a,&b,&c)

#define f0(i,b) for(int i=0;i<(b);i++)
#define f1(i,b) for(int i=1;i<=(b);i++)
#define f2(i,a,b) for(int i=(a);i<=(b);i++)
#define fr(i,b,a) for(int i=(b);i>=(a);i--)

#define CIN ios_base::sync_with_stdio(0); cin.tie(NULL); cout.tie(NULL);
#define TEST_CASE(t) for(int zz=1 ; zz<=t ; zz++)
#define PRINT_CASE printf("Case %d: ",zz)
#define endl            "\n"
#define Debug(x)            cout<<#x " = "<<(x)<<endl
#define NOT_VISITED 0
#define IS_VISITED 1
#define WHITE 0
#define GRAY 1
#define BLACK 2

int fx[4]= {1,-1,0,0};
int fy[4]= {0,0,1,-1};

/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N  (1<<pos);}
//int Reset(int N,int pos){return N= N & ~(1<<pos);}
//bool Check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/

const double PI = acos(-1.0);
const double EPS = 1e-6;
const int MOD = (int)1e9+7;
const int maxn = (int)2e5+5;


typedef long long ll;
const ll inf = 5e18+100;
typedef unsigned long long ull;
typedef pair<ll, int> plli;
typedef pair<int, ll> pill;
typedef vector<int> vi;
int n;

int treemx[4*maxn],treemn[4*maxn], arr[maxn];

void Build(int node, int s, int f){
     if(s==f){
        treemx[node]=arr[f];
        treemn[node]=arr[f];
        return;
    }
    int left=node<<1;
    int right=left+1;
    int mid=(s+f)/2;
    Build(left,s,mid);
    Build(right,mid+1,f);
    treemx[node]=max(treemx[left],treemx[right]);
    treemn[node]=min(treemn[left],treemn[right]);
    return;
}

int querymx(int node, int s, int f, int i, int j){
    if(i>f||j<s){return 0;}
    if(s>=i&&j>=f){return treemx[node];}
    int left= node<<1;
    int right= left+1;
    int mid=(s+f)/2;
    int p1=querymx(left,s,mid,i,j);
    int p2=querymx(right,mid+1,f,i,j);
    return max(p1,p2);
}
int querymn(int node, int s, int f, int i, int j){
    if(i>f||j<s){return MOD;}
    if(s>=i&&j>=f){return treemn[node];}
    int left= node<<1;
    int right= left+1;
    int mid=(s+f)/2;
    int p1=querymn(left,s,mid,i,j);
    int p2=querymn(right,mid+1,f,i,j);
    return min(p1,p2);
}

void solve(){
    cin >> n;

    for(int i=1 ; i<=n ; i++) {
        cin >> arr[i];
    }

    Build(1, 1, n);

    //return;

    bool yes = true;
    for(int i=1 ; i<=n-2 ; i++){
        int mx1 = querymx(1, 1, n, 1, i);
        if(mx1>arr[i+1]) continue;
        int left = i+1, right = n-1, ind = n+1, mid;

        while(left<=right){
            mid = (left+right)/2;
            int mn1 = querymn(1, 1, n, i+1, mid);
            int mx2 = querymx(1, 1, n, mid+1, n);

            if(mn1>mx1){
                left = mid+1;
            } else if(mx2>mx1){
                left = mid+1;
            } else if(mn1<mx1){
                right = mid-1;
            }else if(mx2<mx1){
                right = mid-1;
            }else {
                cout << "YES" << endl;
                cout << i << " " << mid-i << " " << n-mid << endl;
                ind = i;
                break;
            }
        }

        if(ind>n) continue;
        yes = false;
        break;

    }

    if(yes) cout << "NO" << endl;
    return;
}

int main()
{
    CIN

    //cout << "Fi Amanillah" << endl;

    int t = 1;
    cin >> t;

    while(t--){
        solve();
    }

    return 0;
}