Egor has come up with a hard problem for a training camp! Here it is: Given an array $a$ of $n$ positive integers sorted in increasing order, find $4$ indices $i \lt j \lt p \lt q$ such that: $$ a_i\cdot a_q = a_j \cdot a_p $$ He then wrote the checker to this problem: ```cpp // returns true if the solution is found, // returns false if the solution is not found, // makes the verdict Wrong Answer right away if the found solution is not valid bool getAnswer(InStream &stream, vector<long long> a) { string s = stream.readToken("NO|YES"); // PE if the string is not NO or YES if (s == "NO") return false; vector<int> b = stream.readInts(4, 1, (int)a.size()); // 4 indices between 1 and n int i = b[0] - 1, j = b[1] - 1, p = b[2] - 1, q = b[3] - 1; // -1 to make 0-indexed stream.ensuref(i < j && j < p && p < q, "4 indices should be in increasing order"); stream.ensuref(a[q] / a[p] == a[j] / a[i], "the products are not equal"); return true; } ``` The multiplication will overflow `long long`, so Egor used division instead. How smart! Although now Egor might have another problem...

The first line contains one integer $n$ $(4 \le n \le 500\,000)$ — the size of the array. The second line contains the array $a_1, a_2, \dots, a_n$ itself $(1 \le a_1 \lt a_2 \lt \dots \lt a_n \le 10^{18})$.

On the first line print `YES` if there is a solution and print `NO` otherwise. If a solution exists, print the $4$ chosen indices in order $i, j, p, q$, separated by spaces. If there is more than one solution, you can print any one.

The 2023 ICPC Training Camp powered by Huawei. Day 1: Um_nik mod 998 244 353 Contest.