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:
/*
 * @file work2.cpp
 * @path D:\Vscode\code\work2.cpp
 * @author1 low-equipped w_udixixi
 * @author2 Sher丶lock
 * @date  2020-11-25 18:09:59
*/
#pragma GCC optimize(1)
#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<string>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<cmath>
#include<math.h>
#include<iostream>
#include<bitset>
#include<set>
#include<list>
#include<deque>
#include<algorithm>
#include<unordered_map>
#define ll long long
#define ull unsigned long long
#define pb push_back
#define rep(x,a,b) for (int x=a;x<=b;x++)
#define repp(x,a,b) for (int x=a;x<b;x++)
#define db1(x) printf("%d cx\n",x)
#define db2(x,y) printf("%d %d cx\n",x,y)
#define db3(x,y,z) printf("%d %d %d cx\n",x,y,z)
#define db4(x,y,z,q) printf("%d %d %d %d cx\n",x,y,z,q)
#define DB1(x) printf("%lld cx\n",x)
#define DB2(x,y) printf("%lld %lld cx\n",x,y)
#define DB3(x,y,z) printf("%lld %lld %lld cx\n",x,y,z)
#define DB4(x,y,z,q) printf("%lld %lld %lld %lld cx\n",x,y,z,q)
#define W(x) printf("%d\n",x)
#define WW(x) printf("%lld\n",x)
#define pi 3.14159265358979323846
#define mem(a,x) memset(a,x,sizeof a)
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
using namespace std;
const int maxn=2e5+7;
const int INF=0x3f3f3f3f;
const ll INFF=1e18;
struct node
{
    int minn,maxx,l,r;
}tree[maxn<<2];
int t,n,a[maxn];
void pu(int rt)
{
    tree[rt].maxx=max(tree[rt<<1].maxx,tree[rt<<1|1].maxx);
    tree[rt].minn=min(tree[rt<<1].minn,tree[rt<<1|1].minn);
}
void build(int rt,int l,int r)
{
    tree[rt].l=l;
    tree[rt].r=r;
    if (l==r)
    {
        tree[rt].maxx=tree[rt].minn=a[l];
        return;
    }
    int mid=(l+r)>>1;
    build(lson);
    build(rson);
    pu(rt);
}
int quertmin(int L,int R,int rt,int l,int r)
{
    if (L==l&&R==r)return tree[rt].minn;
    int mid=(l+r)>>1;
    if (L>mid)return quertmin(L,R,rson);
    else if (R<=mid)return quertmin(L,R,lson);
    else return min(quertmin(L,mid,lson),quertmin(mid+1,R,rson));
}
int quertmax(int L,int R,int rt,int l,int r)
{
    if (L==l&&R==r)return tree[rt].maxx;
    int mid=(l+r)>>1;
    if (L>mid)return quertmax(L,R,rson);
    else if (R<=mid)return quertmax(L,R,lson);
    else return max(quertmax(L,mid,lson),quertmax(mid+1,R,rson));
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        rep(i,1,n)scanf("%d",&a[i]);
        build(1,1,n);
        bool mark=false;
        rep(i,1,n-2)
        {
            int l=i+1,r=n-1;
            int x=quertmax(1,i,1,1,n);
            while(l<=r)
            {
                int mid=(l+r)>>1;
                int y=quertmin(i+1,mid,1,1,n);
                int z=quertmax(mid+1,n,1,1,n);
                if (x<y||x<z)l=mid+1;
                else if (x>y||x>z)r=mid-1;
                else
                {
                    printf("YES\n");
                    printf("%d %d %d\n",i,mid-i,n-mid);
                    mark=true;
                    break;
                }
            }
            if (mark)break;
        }
        if (!mark)printf("NO\n");
    }
    return 0;
}