It was the third month of remote learning, Nastya got sick of staying at dormitory, so she decided to return to her hometown. In order to make her trip more entertaining, one of Nastya's friend presented her an integer array a. 

Several hours after starting her journey home Nastya remembered about the present. To entertain herself she decided to check, are there four different indices x, y, z, w such that a_x + a_y = a_z + a_w.

Her train has already arrived the destination, but she still hasn't found the answer. Can you help her unravel the mystery?

Input

The first line contains the single integer n (4 ≤ n ≤ 200 000) — the size of the array.

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 2.5 ⋅ 10^6).

Output

Print "YES" if there are such four indices, and "NO" otherwise.

If such indices exist, print these indices x, y, z and w (1 ≤ x, y, z, w ≤ n).

If there are multiple answers, print any of them.

Examples

Input


6
2 1 5 2 7 4


Output


YES
2 3 1 6 

Input


5
1 3 1 9 20


Output


NO

Note

In the first example a_2 + a_3 = 1 + 5 = 2 + 4 = a_1 + a_6. Note that there are other answer, for example, 2 3 4 6.

In the second example, we can't choose four indices. The answer 1 2 2 3 is wrong, because indices should be different, despite that a_1 + a_2 = 1 + 3 = 3 + 1 = a_2 + a_3
//#include<math.h>
#include<algorithm>
#include<stdlib.h>
#include<time.h>
#include<stdio.h>
#include<string.h>
#define un unsigned
#define srd srand(time(0))
#define ll long long
#define con continue
#define gtc getchar()
#define ptc putchar
#define dou double
#define eps 0.00000000001
#define opr operator
#define cl(x,a) memset(x,a,sizeof(x))
#define fo0(i,k) for(i=fr[k];i;i=nx[i])
#define fo1(i,l,r) for(i=l;i<=r;i++)
#define fo2(i,l,r) for(i=l;i>=r;i--)
#define fo(i,n) for(i=1;i<=n;i++)
#define ret return
#define x first
#define cint const int
#define y second
#define opi(x) freopen(x,"r",stdin)
#define opo(x) freopen(x,"w",stdout)
#define tpl template<class T>
#define priq priority_queue
#define mp make_pair
#define use using namespace
#define WT while(T--)
#define pb push_back
#define sz size()
use std;
typedef pair<int,int> pii;typedef pair<int,ll> pil;typedef pair<ll,int> pli;typedef pair<ll,ll> pll;
namespace io
{
	void _(int &k){char c;int e=1;k=0;while((c=gtc)>'9'||c<'0')if(c=='-')e=-1;k=c-'0';while((c=gtc)<='9'&&c>='0'){k*=10;k+=c-'0';}k*=e;}
	void _(ll &k){char c;int e=1;k=0;while((c=gtc)>'9'||c<'0')if(c=='-')e=-1;k=c-'0';while((c=gtc)<='9'&&c>='0'){k*=10;k+=c-'0';}k*=e;}
	void _(char &c){while((c=gtc)==' '||c=='\n');}void _(dou &c){scanf("%lf",&c);}void _(char *s){char c;while((c=gtc)!=EOF&&c!=' '&&c!=10)*s++=c;}
	template<class t1,class t2>void _(t1 &a,t2 &b){_(a);_(b);}template<class t1,class t2,class t3>void _(t1 &a,t2 &b,t3 &c){_(a);_(b);_(c);}
	template<class t1,class t2,class t3,class t4>void _(t1 &a,t2 &b,t3 &c,t4 &d){_(a);_(b);_(c);_(d);}
	template<class t1,class t2,class t3,class t4,class t5>void _(t1 &a,t2 &b,t3 &c,t4 &d,t5 &e){_(a);_(b);_(c);_(d);_(e);}
	void _p(dou k){printf("%.6lf",k);}void _p(char *c){for(;*c;ptc(*c++));}void _p(const char *c){for(;*c;ptc(*c++));}void _p(char c){ptc(c);}
	tpl void _p0(T k){if(k>=10)_p0(k/10);ptc(k%10+'0');}tpl void _p(T k){if(k<0){ptc('-');_p0(-k);}else _p0(k);}tpl void __p(T k){_p(k);ptc(' ');}
	tpl void _pn(T k){_p(k);ptc('\n');}template<class t1,class t2>void _p(t1 a,t2 b){__p(a);_pn(b);}
	template<class t1,class t2,class t3>void _p(t1 a,t2 b,t3 c){__p(a);__p(b);_pn(c);}
	template<class t1,class t2,class t3,class t4>void _p(t1 a,t2 b,t3 c,t4 d){__p(a);__p(b);__p(c);_pn(d);}
	tpl void op(T *a,int n){int i;n--;fo(i,n)__p(a[i]);_pn(a[n+1]);}int gi(){int x;_(x);ret x;}ll gll(){ll x;_(x);ret x;}
}
int gcd(int a,int b){ret b?gcd(b,a%b):a;}void fcl(){fclose(stdin);fclose(stdout);}
void fop(const char *s){char c[256],d[256];cl(c,0);cl(d,0);strcpy(c,s);strcpy(d,s);opi(strcat(c,".in"));opo(strcat(d,".out"));}
int eq(dou a,dou b){return a+eps>=b&&b+eps>=a;}tpl void _ma(T &a,T b){if(a<b)a=b;}tpl void _mi(T &a,T b){if(a>b)a=b;}
cint N=5555555,EE=100000000,GG=1000000000,ima=2147483647;
use io;
int n,m=5000000,a[N],f[N],g[N],T,t3;
void solve(int sm)
{
	int i,j;
	puts("YES");
	fo(i,n)
		if(!a[i])
		{
			a[i]=t3;
			break;
		}
	fo(i,n)
		fo(j,i-1)
			if(a[i]&&a[j]&&a[i]+a[j]==sm)
			{
				__p(i);
				__p(j);
				a[i]=a[j]=0;
				fo(i,n)
					fo(j,i-1)
						if(a[i]&&a[j]&&a[i]+a[j]==sm)
						{
							_p(i,j);
							exit(0);
						}
			}
}
int main()
{
	int i,j,a1,a2,t=0,t1=0,t2=0;
	_(n);
	_mi(n,6000);
	fo(i,n)
	{
		_(a[i]);
		f[a[i]]++;
		if(f[a[i]]==2)
		{
			t1+=a[i];
			t++;
		}
		if(f[a[i]]==4)
		{
			puts("YES");
			fo(j,i-1)
			{
				if(a[j]==a[i])
					__p(j);
			}
			_pn(i);
			return 0;
		}
		if(t==2)
		{
			int e=0;
			puts("YES");
			fo(j,i-1)
			{
				if(a[j]==a[i])
					__p(j);
			}
			for(j=1;j<i;j++)
			{
				if(a[j]==t1-a[i])
				{
					__p(j);
					e++;
					if(e==2)
						break;
				}
			}
			_pn(i);
			return 0;
		}
	}
	fo(i,m)
	{
		if(f[i]>=2)
		{
			t3=i;
			g[i+i]++;
			fo(j,n)
				if(a[j]==t3)
					a[j]=0;
			fo(j,n)
				if(!a[j])
				{
					a[j]=t3;
					break;
				}
		}
	}
	fo(i,n)
		fo(j,i-1)
			if(a[i]&&a[j])
			{
				g[a[i]+a[j]]++;
				if(g[a[i]+a[j]]>=2)
				{
					solve(a[i]+a[j]);
				}
			}
	puts("NO");
}


