A. Hilbert's Hotel

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

Note

In the first test case, every guest is shifted by 1414 rooms, so the assignment is still unique.

In the second test case, even guests move to the right by 11 room, and odd guests move to the left by 11 room. We can show that the assignment is still unique.

In the third test case, every fourth guest moves to the right by 11 room, and the other guests move to the right by 55 rooms. We can show that the assignment is still unique.

In the fourth test case, guests 00 and 11 are both assigned to room 33.

In the fifth test case, guests 11 and 22 are both assigned to room 22.

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <deque>
#include <bitset>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>
//#include <xfunctional>
#define ll              long long
#define PII             pair<int, int>
#define rep(i,a,b)      for(int  i=a;i<=b;i++)
#define dec(i,a,b)      for(int  i=a;i>=b;i--)
#define pb              push_back
#define mk              make_pair
using namespace std;
int dir[4][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979;
const int mod = 998244353;
const int N = 1e6+5;
//if(x<0 || x>=r || y<0 || y>=c)

inline ll read()
{
    ll x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}
ll gcd(ll m, ll n)
{
    return n == 0 ? m : gcd(n, m % n);
}
ll lcm(ll m, ll n)
{
    return m*n / gcd(m, n);
}
int main()
{
    int T;
    cin >> T;
    while (T--)
    {
        int n;
        cin >> n;
        vector<int> a(n),b(n,0);
        for (int i = 0; i < n; i++)
        {
            cin >> a[i];
        }
        for (int i = 0; i < n; i++)
        {
            b[((i + a[i]) % n + n) % n]++;
        }
        int fg = 0;
        for (int i = 0; i < n; i++)
        {
            if (b[i] != 1)
                fg = 1;
        }
        if (fg)
            cout << "NO" << endl;
        else
            cout << "YES" << endl;
    }
    return 0;
}