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
#ifdef SAM
#include "stdc++.h"
#else
#include <bits/stdc++.h>
#endif

using namespace std;

typedef long long ll;
typedef double db;
typedef vector<string> VS;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<ll> VLL; 
typedef vector<VLL> VVLL;
typedef vector<VVLL> VVVLL;
typedef pair<ll,ll> PLL;
typedef map<ll,ll> MLL;
typedef set<ll> SLL;
typedef queue<ll> QLL;

#define rep(x,l,u) for(ll x = l; x < u; x++)
#define rrep(x,l,u) for(ll x = l; x >= u; x--)
#define fe(x,a) for (auto x : a)
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define mst(x,v) memset(x, v, sizeof(x))
#define sz(x) (ll)x.size()
#define lg(x) (ll)x.length()
#define fi first
#define se second
#define pb push_back

#define umap unordered_map
#define uset unordered_set
#define mset multiset
#define pqueue priority_queue

struct DEBUG {
	string open = "[", close = "]", sep = ", ";

	string f(string s) { return s; }
	string f(char c) { return string(1, c); }
	string f(int x) { return to_string(x); }
	string f(ll x) { return to_string(x); }
	string f(db x) { return to_string(x); }
	string f(const char* a) { return a; }
	template <typename A, typename B> string f(pair<A,B> p) { return open + f(p.first) + sep + f(p.second) + close; }
	template <typename A> string f(A v) { string s = open; for (auto a : v) { if (s != open) { s += sep; } s += f(a); } s += close; return s; } 

	void show() {cout << endl;}
	template <typename Head, typename... Tail> void show(Head H, Tail... T) { cout << " " << f(H); show(T...); }
};

#ifdef SAM
#define debug(...) do { DEBUG deb; cout << "[" << #__VA_ARGS__ << "]:", deb.show(__VA_ARGS__); } while (false)
#else
#define debug(...) do {} while (false)
#endif

VLL di = {0,0,1,-1,1,-1,1,-1};
VLL dj = {1,-1,0,0,-1,-1,1,1};

void readstr(string & s, const ll len=400005) { char buf[len]; ll a = scanf("%s", buf); a++; s = buf; }
ll ob(ll i, ll n) { return i < 0 || i >= n; } // out of bounds
ll tp(ll x) { return (1LL<<x); }
ll roundup(ll a, ll b) { return a % b ? a/b + 1 : a/b; }
vector<string> ssplit(string s) { vector<string> ans; stringstream ss; ss << s; while (ss >> s) ans.pb(s); return ans; }
void makemod(ll & x, ll m) { x %= m; x += m; x %= m; }
ll powmod(ll a, ll b, ll m) { if (b == 0) return 1;	ll h = powmod(a, b/2, m); ll ans = h*h%m; return b%2 ? ans*a%m : ans; }
bool isvowel(char ch) { return (ch=='a' || ch=='e' || ch=='i' || ch=='o' || ch=='u'); }
ll sign(ll x) {	return x > 0 ? 1 : x < 0 ? -1 : 0; }

template<typename A, typename B> void upmin(A & x, B v) { x = min(x, (A)v); }
template<typename A, typename B> void upmax(A & x, B v) { x = max(x, (A)v); }

const VLL mods = {(ll)1e9 + 7, 998244353, (ll)1e6 + 3, (ll)1e18 + 5};
const ll mod = mods[0];
const ll inf = mods[3];
const db eps = 1e-10;
const db pi = acos(0) * 2;
const string nl = "\n";

VVLL A;

void yes(VLL inds) {
	sort(all(inds));
	cout << "YES" << endl;
	VLL res = {inds[0], inds[3], inds[2], inds[1]};
	fe(i, res) {
		cout << A[i][1] + 1 << ' ';
	}
	cout << endl;
	exit(0);
}

void no() {
	cout << "NO" << endl;
	exit(0);
}

void solve() {
	ll n;
	cin >> n;
	A.resize(n, VLL(2));
	rep(i,0,n) {
		cin >> A[i][0];
		A[i][1] = i;
	}
	sort(all(A));

	umap<ll, PLL> diffs;
	rep(i,1,n) {
		ll d = A[i][0] - A[i-1][0];
		if (diffs.find(d) != diffs.end()) {
			yes({diffs[d].first, diffs[d].second, i, i-1});
			// be gone
		}

		if (i > 1) {
			diffs[A[i-1][0] - A[i-2][0]] = {i-1, i-2};
		}
	}

	// n should be less than 2000 now
	diffs.clear();

	rep(i,0,n) {
		rep(j,i+1,n) {
			ll s = A[j][0] - A[i][0];
			if (diffs.find(s) != diffs.end()) {
				yes({i, j, diffs[s].first, diffs[s].second});
			}
		}

		rep(j,0,i) {
			diffs[A[i][0] - A[j][0]] = {i, j};
		}
	}
	no();
}

signed main() {
	ll t = 1;
	// cin >> t;
	rep(i,0,t) solve();
	return 0;
}