Problem: 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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;

///#include<ext/pb_ds/assoc_container.hpp>
///#include<ext/pb_ds/tree_policy.hpp>
///using namespace __gnu_pbds;
///template<typename T> using orderset = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
///(X).order_of_key(value) /// return lower_bound(value)
///(*X).find_by_order(index) /// return value (0 index)

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); /// mt19937_64 (long long)

typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
typedef vector<int> vii;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<vii> vvii;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<string,int> psi;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef vector<bool> vbb;
typedef vector<string> vss;

const int N = 5000001;
const int M = 20;
const int MOD = 1000000007;
const int inf = 1234567890;
const ll INF = 1122334455667788990;

#define fast() ios_base::sync_with_stdio(false),cin.tie(NULL)
#define next(x) next_permutation(all(x))
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define rev(a) reverse(all(a))
#define eb emplace_back
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define B begin()
#define E end()
#define ub upper_bound
#define lb lower_bound
#define yes cout<<"YES"<<nl
#define no cout<<"NO"<<nl
#define error cout<<-1<<nl
#define space ' '
#define nl '\n'
#define PI acos(-1)
#define sqr(x) ((x)*(x))
#define MAX(x) *max_element(all(x))
#define MIN(x) *min_element(all(x))
#define input freopen("input.txt","r",stdin)
#define output freopen("output.txt","w",stdout)
#define Case(x) cout<<"Case "<<x<<": "
#define strtoint(a) atoi(a.c_str())
#define dbg(x) cerr<<#x<<" is "<<x<<'\n';
#define fix(x) cout<<fixed<<setprecision(x)
#define leastonepos(x) __builtin_ffs(x)
#define leadingoff(x) __builtin_clz(x)
#define trailingoff(x) __builtin_ctz(x)
#define numofone(x) __builtin_popcountll(x)
#define time() cerr<<"Time elapsed : "<<clock()*1000.0/CLOCKS_PER_SEC<<"ms"<<'\n'

#define coutv(v) for(auto it:v)cout<<it<<' ';cout<<nl;
#define cinv(v) for(auto &it:v)cin>>it;

///................Graph's Move.................
///const int dx[] = {+1,-1,+0,+0}; ///Rock's Move
///const int dy[] = {+0,+0,+1,-1}; ///Rock's Move
///const int dx[] = {+0,+0,+1,-1,-1,+1,-1,+1}; ///King's Move
///const int dy[] = {-1,+1,+0,+0,+1,+1,-1,-1}; ///king's Move
///const int dx[] = {-2,-2,-1,-1,+1,+1,+2,+2}; ///knight's Move
///const int dy[] = {-1,+1,-2,+2,-2,+2,-1,+1}; ///knight's Move
///*.....................-_-........................*///




int main()
{
    fast();
    int tc=1,cs=0;
    ///cin>>tc;
    while(tc--)
    {
        int n,i,j;
        cin>>n;
        vii v(n);
        cinv(v);
        vii l(N),r(N);
        for(i=0; i<n; i++)
            for(j=0; j<n; j++)
            {
                if(j<=i)continue;
                int s=v[i]+v[j];
                if(r[s])
                {
                    if(l[s]!=i&&l[s]!=j&&r[s]!=i&&r[s]!=j)
                    {
                        yes;
                        cout<<l[s]+1<<space<<r[s]+1<<space<<i+1<<space<<j+1<<nl;
                        return 0;
                    }
                }
                else if(!r[s])
                {
                    l[s]=i;
                    r[s]=j;
                }
            }
        no;
    }
}