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
// Mivik 2021.4.7
#include <cstdio>
#include <cstring>
#include <iostream>

#define d isdigit(c)
#define g c=t.get();
#define L return
#define K break
#define A(c,a,b)if(c)a else b;
#define I(c,a)if(c)a;
#define Y goto E
struct MI{private:char bb[1 << 14];FILE*f;char*bs,*be;char e;bool o,l;public:MI():f(stdin),bs(0),be(0){}
#ifdef __linux__
#endif
inline char get(){if(o){o=0;L e;}
#ifdef MIVIK
char c=fgetc(f);I(c==-1,l=1)L c;
#else
I(bs==be,be=(bs=bb)+fread(bb,1,sizeof(bb),f))I(bs==be,{l=1;L-1;})L*bs++;
#endif
}inline void unget(char c){o=1;e=c;}template<class T>inline T read(){T r;*this>r;L r;}template<class T>inline MI&operator>(T&);};template<class T>struct Q{const static bool U=T(-1)>=T(0);inline void operator()(MI&t,T&r)const{r=0;char c;bool y=0;A(U,for(;;){g I(c==-1,Y)I(d,K)},for(;;){g I(c==-1,Y)A(c=='-',{g I(d,{y=1;K;})},I(d,K))})for(;;){I(c==-1,Y)A(d,r=r*10+(c^48);,K)g}t.unget(c);E:;I(y,r=-r)}};template<>struct Q<char>{};template<class T>inline MI&MI::operator>(T&t){Q<T>()(*this,t);L*this;}
#undef d
#undef g
#undef L
#undef K
#undef A
#undef I
#undef Y
template<class T>std::ostream& operator<(std::ostream&out,const T&t){return out<<t;}
#define endl ('\n')
#define P(x) cout<#x" = "<(x)<endl
#define R (cin.read<int>())

using std::cout;

#include <algorithm>
#include <cassert>
#include <vector>

MI cin;

typedef long long qe;

const int mod = 998244353;

template<typename T> inline bool gmax(T &a, const T &b) { return a < b && (a = b, true); }

inline int add(int x, int y) { return (x += y) >= mod ? x - mod : x; }
inline int sub(int x, int y) { return (x -= y) < 0 ? x + mod : x; }
inline qe ksm(qe x, int p = mod - 2) {
	qe ret = 1;
	for (; p; p >>= 1, (x *= x) %= mod)
		if (p & 1) (ret *= x) %= mod;
	return ret;
}
std::vector<int> rt, r, inv;
inline void init_ntt(int N) {
	rt.resize(1 << N); r.resize(1 << N);
	inv.resize(N + 1);
	for (int i = 0, j = 1, q = 0; i <= N; ++i, j = (q = j) << 1) {
		inv[i] = ksm(j);
		const int del = ksm(3, (mod - 1) / j);
		for (int k = 0, c = 1; k < q; ++k, c = (qe)c * del % mod) rt[q | k] = c;
	}
}
inline void ntt(int *v, int len, bool rev) {
	static int last;
	if (last != len) {
		for (int i = 1; i < len; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1)? (len >> 1): 0);
		last = len;
	}
	for (int i = 1; i < len; ++i) if (i < r[i]) std::swap(v[i], v[r[i]]);
	assert(!(len & (len - 1)));
	for (int i = 1, q = 2; i < len; i = q, q <<= 1)
	for (int j = 0; j < len; j += q)
	for (int k = 0; k < i; ++k) {
		const int x = v[j | k], y = (qe)v[i | j | k] * rt[i | k] % mod;
		v[j | k] = add(x, y);
		v[i | j | k] = sub(x, y);
	}
	if (!rev) return; std::reverse(v + 1, v + len);
	for (int i = 0, r = inv[__builtin_ctz(len)]; i < len; ++i) v[i] = (qe)v[i] * r % mod;
}
struct poly : public std::vector<int> {
#define v (*this)
	static inline int round_up(int x) { return (x & (x - 1))? (1 << (32 - __builtin_clz(x))): x; }

	poly() {}
	inline poly& ntt(int len, bool rev) { resize(round_up(len), 0); ::ntt(data(), size(), rev); return v; }
	inline poly& ntt(bool rev) { return ntt(size(), rev); }
	inline poly& imul(const poly &a, int l) {
		static poly tmp;
		if (empty() || a.empty()) return clear(), v;
		const int len = round_up(l);
		(tmp = a).ntt(len, 0); ntt(len, 0);
		for (int i = 0; i < len; ++i) v[i] = (qe)v[i] * tmp[i] % mod;
		return ntt(len, 1), resize(l), v;
	}
	inline poly& operator*=(const poly &a) { return imul(a, size() + a.size() - 1); }
#undef v
};
template<> struct Q<poly> {  };

const int $n = 200005;
const int $v = 5000005;

int n, v[$n], cnt[$v], lst[$v], pre[$n], tar[$n];
poly F;
inline void link(int x, int y) {
	static int et;
	pre[++et] = lst[x];
	lst[x] = et;
	tar[et] = y;
}
inline void print(int a, int b, int c, int d) {
	static int t[4];
	t[0] = a;
	t[1] = b;
	t[2] = c;
	t[3] = d;
	std::sort(t, t + 4);
	assert(t[0] != t[1] && t[1] != t[2] && t[2] != t[3]);
	assert(v[a] + v[b] == v[c] + v[d]);
	cout < "YES\n";
	cout < a < ' ' < b < ' ' < c < ' ' < d < endl;
}
inline void find(int a, int b, int c, int d) {
	int na, nb, nc, nd;
	na = tar[lst[a]];
	lst[a] = pre[lst[a]];
	nb = tar[lst[b]];
	lst[b] = pre[lst[b]];
	nc = tar[lst[c]];
	lst[c] = pre[lst[c]];
	nd = tar[lst[d]];
	lst[d] = pre[lst[d]];
	print(na, nb, nc, nd);
}
int main() {
	cin > n;
	int c = 0, mx = 0, t1, t2;
	for (int i = 1; i <= n; ++i) {
		const int v = ::v[i] = R;
		link(v, i);
		if (++cnt[v] == 2) {
			if (++c == 1) {
				t1 = tar[pre[lst[v]]];
				t2 = i;
			} else
				return print(t1, tar[pre[lst[v]]], t2, i), 0;
		} else if (cnt[v] == 4)
			return find(v, v, v, v), 0;
		gmax(mx, v);
	}
	for (int i = 0; i <= mx; ++i) {
		if (cnt[i] < 2) continue;
		const int w = i * 2;
		for (int j = 0; j <= w && j <= mx; ++j) {
			if (j == i || !cnt[j] || !cnt[w - j]) continue;
			return find(i, i, j, w - j), 0;
		}
	}
	init_ntt(32 - __builtin_clz(mx * 2 + 1));
	F.resize(mx + 1);
	for (int i = 0; i <= mx; ++i) F[i] = !!cnt[i];
	F.ntt(F.size() * 2 - 1, 0);
	for (int &v : F) v = (qe)v * v % mod;
	F.ntt(1);
	t1 = -1;
	for (int i = 0; i < F.size(); ++i) {
		if (F[i] < 4) continue;
		for (int j = 0; j <= i; ++j) {
			if (!lst[j] || !lst[i - j]) continue;
			if (t1 != -1)
				return find(t1, t2, j, i - j), 0;
			else {
				t1 = j;
				t2 = i - j;
			}
		}
		assert(0);
	}
	cout < "NO\n";
}