Codeforces Round #639 (Div. 2) C. Hilbert's Hotel（数学）

Hilbert's Hotel is a very unusual hotel since the number of rooms is infinite! In fact, there is exactly one room for every integer, including zero and negative integers. Even stranger, the hotel is currently at full capacity, meaning there is exactly one guest in every room. The hotel's manager, David Hilbert himself, decides he wants to shuffle the guests around because he thinks this will create a vacancy (a room without a guest).

For any integer kk and positive integer nn , let kmodnkmodn denote the remainder when kk is divided by nn . More formally, r=kmodnr=kmodn is the smallest non-negative integer such that k−rk−r is divisible by nn . It always holds that 0≤kmodn≤n−10≤kmodn≤n−1 . For example, 100mod12=4100mod12=4 and (−1337)mod3=1(−1337)mod3=1 .

Then the shuffling works as follows. There is an array of nn integers a0,a1,…,an−1a0,a1,…,an−1 . Then for each integer kk , the guest in room kk is moved to room number k+akmodnk+akmodn .

After this shuffling process, determine if there is still exactly one guest assigned to each room. That is, there are no vacancies or rooms with multiple guests.

Input

Each test consists of multiple test cases. The first line contains a single integer tt (1≤t≤1041≤t≤104 ) — the number of test cases. Next 2t2t lines contain descriptions of test cases.

The first line of each test case contains a single integer nn (1≤n≤2⋅1051≤n≤2⋅105 ) — the length of the array.

The second line of each test case contains nn integers a0,a1,…,an−1a0,a1,…,an−1 (−109≤ai≤109−109≤ai≤109 ).

It is guaranteed that the sum of nn over all test cases does not exceed 2⋅1052⋅105 .

Output

For each test case, output a single line containing "YES" if there is exactly one guest assigned to each room after the shuffling process, or "NO" otherwise. You can print each letter in any case (upper or lower).

Example

Input

Copy

6
1
14
2
1 -1
4
5 5 5 1
3
3 2 1
2
0 1
5
-239 -2 -100 -3 -11

Output

Copy

YES
YES
YES
NO
NO
YES
就是判断这个序列的每个数a[i]加上其对应位置的i(i:1~n)后能否构成模n的一个完全剩余系（是这么说的吧
注意负数的话可以加上一个很大的n的倍数再模。
证明的话可以参考https://www.cnblogs.com/iss-ue/p/12844330.html
代码用set瞎写的

#include <bits/stdc++.h>
using namespace std;
long long n,a[200005];
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        cin>>n;
        int i;
        set<long long>s;
        for(i=1;i<=n;i++)
        {
            scanf("%lld",&a[i]);
            a[i]+=i;
            s.insert((a[i]+n*20000000000)%n);
        }
        if(s.size()==n)cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
    return 0;
}