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 k and positive integer n, let kmod n denote the remainder when k is divided by n. More formally, r=kmod n is the smallest non-negative integer such that k-r is divisible by n. It always holds that 0≤ kmod n≤ n-1. For example, 100mod 12=4 and (-1337)mod 3=1.

Then the shuffling works as follows. There is an array of n integers a_0,a_1,…,a_{n-1}. Then for each integer k, the guest in room k is moved to room number k+a_{kmod n}.

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 t (1≤ t≤ 10^4) — the number of test cases. Next 2t lines contain descriptions of test cases.

The first line of each test case contains a single integer n (1≤ n≤ 2⋅ 10^5) — the length of the array.

The second line of each test case contains n integers a_0,a_1,…,a_{n-1} (-10^9≤ a_i≤ 10^9).

It is guaranteed that the sum of n over all test cases does not exceed 2⋅ 10^5.

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


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


YES
YES
YES
NO
NO
YES

Note

In the first test case, every guest is shifted by 14 rooms, so the assignment is still unique.

In the second test case, even guests move to the right by 1 room, and odd guests move to the left by 1 room. We can show that the assignment is still unique.

In the third test case, every fourth guest moves to the right by 1 room, and the other guests move to the right by 5 rooms. We can show that the assignment is still unique.

In the fourth test case, guests 0 and 1 are both assigned to room 3.

In the fifth test case, guests 1 and 2 are both assigned to room 2.
//x<<y=x*2^y,x>>y=x/2^y
//1e5+3 is prime
// in a matrix, all elements of a primary diagonal have constant diff of coordinates
// and a secondary diagonal has constant sum of coordinates
//use ll() for using an integer in self-built fn
//(x&-x) bitwise and of (x and its 2's complement) returns (last set bit)
//eg if x=1010 then it will return 0010 
// careful dont print empty container(run time error) 
//v.erase O(n)
//use ("\n") instead of endl as endl flushes the output buffer
//every time so takes more time than \n (TLE)
//stoll() and to_string((less than 1e19))
//INT_MAX is approx 3e10
//For sets use set_name.lower_bound(x)(strictly O(logn)) 
//NOT lb(all(s),x)(O(logn) for random access (eg vector) but for sets it is O(n));
#include<bits/stdc++.h>
typedef long long int ll;
#define ull unsigned long long int
#define lld long double
#define endl "\n"
#define fi first 
#define sec second 
#define sp setprecision 
#define lb lower_bound
#define ub upper_bound
#define For(i, a, b) for(long long int i = (a); i <= (b); i++)
#define Forr(i, a, b) for(long long int i = (a); i >= (b); i--)
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define vll vector<ll>
#define pll pair<ll,ll>
#define vlld vector<lld>
#define vi vector<int>
#define vch vector<char>
#define sll set<ll>
#define sch set<ch>
#define vpll vector< pair<ll,ll> >
#define vpii vector< pair<int,int> >
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
#define mll map<ll, ll>
#define mcll map<char,ll>
#define sz(container) ll((container).size())
#define fill(a,b) memset(a, b, sizeof(a))
#define fast_io ios_base::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL)
using namespace std;
lld pi=3.1415926535897932;
const ll mod=1000000007;
const ll MOD=998244353;
const ll dx[4]={-1, 0, 1, 0} , dy[4]={0, 1, 0, -1};
const ll dxx[8]={-1, -1, 0, 1, 1, 1, 0, -1} , dyy[8]={0, 1, 1, 1, 0, -1, -1, -1};
string ds="RLDU";
/***************************************
struct cmp 
{
    bool operator() (const pll &a, const pll &b)  const 
    {
        ll lena = a.sec - a.fi + 1;
        ll lenb = b.sec - b.fi + 1;
        if (lena == lenb) return a.fi < b.fi;
        return lena > lenb;
    }
};
********************************/


ll lcm(ll a, ll b)
{
    ll g=__gcd(a, b);
    return a/g*b;
}

ll binpow(ll a, ll b,ll m) 
{
    a %= m;
    ll res = 1;
    while (b > 0) 
    {
        if (b & 1)
            res = res * a % m;
        a = a * a % m;
        b >>= 1;
    }
    return res;
}
ll modinv(ll n,ll mod) 
{
    return binpow(n, mod - 2,mod);
}


/**************coding****************************************/
int main()
{

    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif
    fast_io;
    ll T=1;cin>>T;
    For(i,1,T)
    {
        ll n;cin>>n;
        vll a(n);
        For(j,0,n-1) cin>>a[j];
        sll temp;
        For(j,0,n-1)
        {
            temp.insert((j+a[j])%n);
        }
        if(sz(temp)==n) cout<<"YES";
        else cout<<"NO";
        cout<<endl;
    } 
}