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.
/**
 *  “Experience is the name everyone gives to their mistakes.” – Oscar Wilde
 *
 *  author  :   prodipdatta7
 *  created :   Wednesday 06-May, 2020  09:12:33 PM
**/

//#include <bits/stdc++.h>
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <stack>
#include <queue>
#include <deque>
#include <iterator>
#include <bitset>
#include <assert.h>
#include <new>
#include <sstream>
#include <time.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx,avx2,fma")
// #pragma GCC optimize("unroll-loops")

using namespace std ;
using namespace __gnu_pbds ;

#define     ll          long long
#define     ld          long double
#define     ull         unsigned long long
#define     pii         pair<int,int>
#define     pll         pair<ll,ll>
#define     vi          vector<int>
#define     vll         vector<ll>
#define     vvi         vector<vector<int>>
#define     debug(x)    cerr << #x << " = " << x << '\n' ;
#define     rep(i,b,e)  for(__typeof(e) i = (b) ; i != (e + 1) - 2 * ((b) > (e))  ; i += 1 - 2 * ((b) > (e)))
#define     all(x)      x.begin() , x.end()
#define     rall(x)     x.rbegin() , x.rend()
#define     sz(x)       (int)x.size()
#define     ff          first
#define     ss          second
#define     pb          push_back
#define     eb          emplace_back
#define     mem(a)      memset(a , 0 ,sizeof a)
#define     memn(a)     memset(a , -1 ,sizeof a)
#define     fread       freopen("input.txt","r",stdin)
#define     fwrite      freopen("output.txt","w",stdout)
#define     TN          typename
#define     FastIO      ios_base::sync_with_stdio(false) ; cin.tie(NULL) ; cout.tie(NULL) ;

typedef tree <int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update > ordered_set;
typedef tree <int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update > ordered_multiset;

/*
Note :  There is a problem with erase() function in ordered_multiset (for less_equal<int> tag).
        lower_bound() works as upper_bound() and vice-versa.
        Be careful to use.
        i) find_by_order(k) : kth smallest element counting from 0 .
        ii) order_of_key(k) : number of elements strictly smaller than k.
*/

/*###############-> Input Section <-###############*/

template <TN T> inline void Int(T &a) {
    bool minus = false; a = 0; char ch = getchar();
    while (true) { if (ch == '-' or (ch >= '0' && ch <= '9')) break; ch = getchar(); }
    if (ch == '-') minus = true; else a = ch - '0';
    while (true) { ch = getchar(); if (ch < '0' || ch > '9') break; a = a * 10 + (ch - '0'); }
    if (minus)a *= -1 ;
}
template < TN T, TN T1 > inline void Int(T &a, T1 &b) {Int(a), Int(b) ;}
template < TN T, TN T1, TN T2 > inline void Int(T &a, T1 &b, T2 &c) {Int(a, b), Int(c) ;}
template < TN T, TN T1, TN T2, TN T3 > inline void Int(T &a, T1 &b, T2 &c, T3 &d) {Int(a, b), Int(c, d) ;}
template < TN T, TN T1, TN T2, TN T3, TN T4> inline void Int(T &a, T1 &b, T2 &c, T3 &d, T4 &e) {Int(a, b), Int(c, d, e) ;}

/*###############-> Debugger <-###############*/

#ifndef ONLINE_JUDGE
#define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
void err(istream_iterator<string> it) {cout << endl ;}
template<TN T, TN... Args>
void err(istream_iterator<string> it, T a, Args... args) {
    cerr << *it << " = " << a << ' ' ;
    err(++it, args...);
}
#else
#define error(args...)
#endif

template<class ValueType>
bool unstable_remove(
    typename std::vector<ValueType>& container,
    typename std::vector<ValueType>::iterator it
) {
    auto lastEl = container.end() - 1;
    if (it != lastEl) {
        *it = std::move(*lastEl);
    }
    container.pop_back();
}

/******************************************
#Rows are numbered from top to bottom.
#Columns are numbered from left to right.
Moves : L, U, R, D, LU, RU, RD, LD.
******************************************/
int dx[8] = {0, -1, 0, 1, -1, -1, 1, 1} ;
int dy[8] = { -1, 0, 1, 0, -1, 1, 1, -1} ;

/*###############-> Constraints <-###############*/

const int N          = (int)2e5 + 5 ;
const int maxN       = (int)1e6 + 6 ;
const ll  Mod        = (ll)1e9 + 7 ;
const int inf        = (int)2e9 ;
const ll  Inf        = (ll)1e18 ;
const int mod        = (int)1e9 + 7 ;


inline int add(int a, int b, int mod) {a += b ; return a >= mod ? a - mod : a ;}
inline int sub(int a, int b, int mod) {a -= b ; return a < 0 ? a + mod : a ;}
inline int mul(int a, int b, int mod) {return (ll)a * b % mod ;}

/*...... ! Code starts from here ! ......*/


int main() {
#ifndef ONLINE_JUDGE
    clock_t tStart = clock();
    freopen("input.txt", "r", stdin) ;
    freopen("output.txt", "w", stdout) ;
#endif

    int test = 1 , tc = 0 ;
    Int(test) ;
    map < int, int > mp ; 
    while (test--) {
        int n ; Int(n) ;
        vi a(n) ; 
        for(int i = 0 ; i < n ; ++i){
            Int(a[i]) ; 
        }
        bool ok = 1 ;
        for(int i = 1 ; i <= n ; ++i){
            int p = i + a[i % n]  ;
            if(mp[p] == 1){
                ok = 0 ;
            }
            mp[p]++ ; 
        }
        puts(ok ? "YES" : "NO") ;
        mp.clear() ; 
    }

#ifndef ONLINE_JUDGE
    fprintf(stderr, "\nRuntime: %.10fs\n", (double) (clock() - tStart) / CLOCKS_PER_SEC);
#endif
    return 0 ;
}