B. Playing with GCD

B. Playing with GCD

You are given an integer array $a$ of length $n$.

Does there exist an array $b$ consisting of $n+1$ positive integers such that $a_i=gcd(b_i,b_{i+1})$ for all $i$ ($1 \leq i \leq n)$?

Note that $gcd(x,y)$ denotes the greatest common divisor (GCD) of integers $x$ and $y$.

Input

Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \leq t \leq {10}^{5}$). Description of the test cases follows.

The first line of each test case contains an integer $n$ ($1 \leq n \leq {10}^{5}$) — the length of the array $a$.

The second line of each test case contains $n$ space-separated integers $a_1,a_2, \dots ,a_n$ representing the array $a$ ($1 \leq a_i \leq {10}^{4}$).

It is guaranteed that the sum of $n$ over all test cases does not exceed ${10}^{5}$.

Output

For each test case, output "$\text{YES}$" if such $b$ exists, otherwise output "$\text{NO}$". You can print each letter in any case (upper or lower).

Example

input

4
1
343
2
4 2
3
4 2 4
4
1 1 1 1

output

YES
YES
NO
YES

Note

In the first test case, we can take $b=[343,343]$.

In the second test case, one possibility for $b$ is $b=[12,8,6]$.

In the third test case, it can be proved that there does not exist any array b that fulfills all the conditions.

 

解题思路

　　比赛的时候想了快一个小时都没想到怎么做。关键就是在于根据已有的数组$a$去构造出数组$b$。

　　对于$a_i$，有$a_{i-1} = gcd(b_{i-1}, b_{i})$，$a_{i} = gcd(b_{i}, b_{i+1})$，可以发现$b_i$既是$a_{i-1}$的倍数，也是$a_i$的倍数，即满足$b_i \, \% \, a_{i-1} = 0, \, b_i \, \% \, a_i = 0$，因此$b_i$就是$a_{i-1}$和$a_{i}$的公倍数。但这里我们取$b_i = lcm(a_{i-1},a_{i})$，这是因为如果$b_i$取最小公倍数的倍数的话，即$k \times lcm(a_{i-1}, a_{i})$，$k$是大于$1$的整数，那么就有$$gcd(b_{i}, b_{i+1}) = gcd(k_{i} \times lcm(a_{i-1},a_{i}), k_{i+1} \times lcm(a_{i},a_{i+1})) ~\geq~ gcd(lcm(a_{i-1},a_{i}), lcm(a_{i},a_{i+1})) = a_{i}$$

　　即对于任意的$b_i$如果不是取$lcm(a_{i-1},a_{i})$，那么经过求gcd操作后得到的不一定是原来的$a$数组（跟确切的意思是，如果每一个$b_i$都满足$b_i = lcm(a_{i-1},a_{i})$，且经过gcd操作后可以得到$a$数组，那么如果$b$的构造方式为$b_i = k_i \times lcm(a_{i-1}, a_{i})$，此时经过gcd操作后就不一定可以得到$a$数组，因为得到的结果可能是原来位置上的数的倍数）。

　　当然，如果$i = 1$，那么$b_1 = a_1$；如果$i = n$，那么$b_{n+1} = a_n$。

　　最后还要判断构造得到的$b$数组经过gcd操作后能否得到$a$数组，如果发现$gcd(b_{i}, b_{i+1}) \ne a_{i}$，那么就不存在解。

　　AC代码如下：

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

const int N = 1e5 + 10;

int a[N], b[N];

int gcd(int a, int b) {
    return b ? gcd(b, a % b) : a;
}

void solve() {
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%d", a + i);
    }
    b[0] = a[0], b[n] = a[n - 1];
    for (int i = 1; i < n; i++) {
        b[i] = a[i] / gcd(a[i], a[i - 1]) * a[i - 1]; // lcm(a[i], a[i - 1])
    }
    for (int i = 0; i < n; i++) {
        if (gcd(b[i], b[i + 1]) != a[i]) {
            printf("NO\n");
            return;
        }
    }
    printf("YES\n");
}

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        solve();
    }
    
    return 0;
}

 

参考资料

　　Codeforces Round #825 (Div. 2) Editorial：https://codeforces.com/contest/1736