## 题目描述

You are given $n$ integers $a_1, a_2, \ldots, a_n$ . Find the maximum value of $max(a_l, a_{l + 1}, \ldots, a_r) \cdot min(a_l, a_{l + 1}, \ldots, a_r)$ over all pairs $(l, r)$ of integers for which $1 \le l < r \le n$ .

## 输入格式

The first line contains a single integer $t$ ( $1 \le t \le 10\,000$ ) — the number of test cases.

The first line of each test case contains a single integer $n$ ( $2 \le n \le 10^5$ ).

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ( $1 \le a_i \le 10^6$ ).

It is guaranteed that the sum of $n$ over all test cases doesn't exceed $3 \cdot 10^5$ .

## 输出格式

For each test case, print a single integer — the maximum possible value of the product from the statement.

## 输入输出样例

### 输入 #1

4
3
2 4 3
4
3 2 3 1
2
69 69
6
719313 273225 402638 473783 804745 323328

### 输出 #1

12
6
4761
381274500335

## 分析

• 第一种情况，当 $b\lt c$ 时，解为 $a\cdot c$
• 第二种情况，当 $a\gt c$ 时，解为 $b\cdot c$

## 代码

#include<bits/stdc++.h>
#define int long long
using namespace std;
int t,n;
int ans;
int a[100001];
signed main(){
scanf("%lld",&t);
while(t--){
memset(a,0,sizeof(a));
ans=0;
scanf("%lld",&n);
for(int i=1;i<=n;i++)scanf("%lld",&a[i]);
for(int i=1;i<=n;i++){
ans=max(ans,a[i]*a[i+1]);
}
printf("%lld\n",ans);
}
}