Problem: Polycarp has 26 tasks. Each task is designated by a capital letter of the Latin alphabet.

The teacher asked Polycarp to solve tasks in the following way: if Polycarp began to solve some task, then he must solve it to the end, without being distracted by another task. After switching to another task, Polycarp cannot return to the previous task.

Polycarp can only solve one task during the day. Every day he wrote down what task he solved. Now the teacher wants to know if Polycarp followed his advice.

For example, if Polycarp solved tasks in the following order: "DDBBCCCBBEZ", then the teacher will see that on the third day Polycarp began to solve the task 'B', then on the fifth day he got distracted and began to solve the task 'C', on the eighth day Polycarp returned to the task 'B'. Other examples of when the teacher is suspicious: "BAB", "AABBCCDDEEBZZ" and "AAAAZAAAAA".

If Polycarp solved the tasks as follows: "FFGZZZY", then the teacher cannot have any suspicions. Please note that Polycarp is not obligated to solve all tasks. Other examples of when the teacher doesn't have any suspicious: "BA", "AFFFCC" and "YYYYY".

Help Polycarp find out if his teacher might be suspicious.

Input

The first line contains an integer t (1 ≤ t ≤ 1000). Then t test cases follow.

The first line of each test case contains one integer n (1 ≤ n ≤ 50) — the number of days during which Polycarp solved tasks.

The second line contains a string of length n, consisting of uppercase Latin letters, which is the order in which Polycarp solved the tasks.

Output

For each test case output: 

  * "YES", if the teacher cannot be suspicious; 
  * "NO", otherwise. 



You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).

Example

Input


5
3
ABA
11
DDBBCCCBBEZ
7
FFGZZZY
1
Z
2
AB


Output


NO
NO
YES
YES
YES

Solution (in CPP):

Solution:
#define LOCAL
#include <bits/stdc++.h>
using namespace std;
#pragma region Macros
typedef long long ll;
typedef __int128_t i128;
typedef unsigned int uint;
typedef unsigned long long ull;
#define ALL(x) (x).begin(), (x).end()

template <typename T> istream& operator>>(istream& is, vector<T>& v) {
    for (T& x : v) is >> x;
    return is;
}
template <typename T> ostream& operator<<(ostream& os, const vector<T>& v) {
    for (int i = 0; i < (int)v.size(); i++) {
        os << v[i] << (i + 1 == (int)v.size() ? "" : " ");
    }
    return os;
}
template <typename T, typename U> ostream& operator<<(ostream& os, const pair<T, U>& p) {
    os << '(' << p.first << ',' << p.second << ')';
    return os;
}
template <typename T, typename U, typename V> ostream& operator<<(ostream& os, const tuple<T, U, V>& t) {
    os << '(' << get<0>(t) << ',' << get<1>(t) << ',' << get<2>(t) << ')';
    return os;
}
template <typename T, typename U, typename V, typename W> ostream& operator<<(ostream& os, const tuple<T, U, V, W>& t) {
    os << '(' << get<0>(t) << ',' << get<1>(t) << ',' << get<2>(t) << ',' << get<3>(t) << ')';
    return os;
}
template <typename T, typename U> ostream& operator<<(ostream& os, const map<T, U>& m) {
    os << '{';
    for (auto itr = m.begin(); itr != m.end();) {
        os << '(' << itr->first << ',' << itr->second << ')';
        if (++itr != m.end()) os << ',';
    }
    os << '}';
    return os;
}
template <typename T, typename U> ostream& operator<<(ostream& os, const unordered_map<T, U>& m) {
    os << '{';
    for (auto itr = m.begin(); itr != m.end();) {
        os << '(' << itr->first << ',' << itr->second << ')';
        if (++itr != m.end()) os << ',';
    }
    os << '}';
    return os;
}
template <typename T> ostream& operator<<(ostream& os, const set<T>& s) {
    os << '{';
    for (auto itr = s.begin(); itr != s.end();) {
        os << *itr;
        if (++itr != s.end()) os << ',';
    }
    os << '}';
    return os;
}
template <typename T> ostream& operator<<(ostream& os, const multiset<T>& s) {
    os << '{';
    for (auto itr = s.begin(); itr != s.end();) {
        os << *itr;
        if (++itr != s.end()) os << ',';
    }
    os << '}';
    return os;
}
template <typename T> ostream& operator<<(ostream& os, const unordered_set<T>& s) {
    os << '{';
    for (auto itr = s.begin(); itr != s.end();) {
        os << *itr;
        if (++itr != s.end()) os << ',';
    }
    os << '}';
    return os;
}
template <typename T> ostream& operator<<(ostream& os, const deque<T>& v) {
    for (int i = 0; i < (int)v.size(); i++) {
        os << v[i] << (i + 1 == (int)v.size() ? "" : " ");
    }
    return os;
}

void debug_out() { cerr << '\n'; }
template <class Head, class... Tail> void debug_out(Head&& head, Tail&&... tail) {
    cerr << head;
    if (sizeof...(Tail) > 0) cerr << ", ";
    debug_out(move(tail)...);
}
#ifdef LOCAL
#define debug(...)                                                                   \
    cerr << " ";                                                                     \
    cerr << #__VA_ARGS__ << " :[" << __LINE__ << ":" << __FUNCTION__ << "]" << '\n'; \
    cerr << " ";                                                                     \
    debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif

template <typename T> T gcd(T x, T y) { return y != 0 ? gcd(y, x % y) : x; }
template <typename T> T lcm(T x, T y) { return x / gcd(x, y) * y; }

template <class T1, class T2> inline bool chmin(T1& a, T2 b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template <class T1, class T2> inline bool chmax(T1& a, T2 b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
#pragma endregion

const int INF = 1e9;
const long long IINF = 1e18;
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
const char dir[4] = {'D', 'R', 'U', 'L'};
const long long MOD = 1000000007;
// const long long MOD = 998244353;

void solve() {
    int n;
    string S;
    cin >> n >> S;
    vector<vector<int>> pos(26);
    for (int i = 0; i < n; i++) pos[S[i] - 'A'].emplace_back(i);

    for (int i = 0; i < 26; i++) {
        for (int j = 1; j < pos[i].size(); j++) {
            if (pos[i][j] - pos[i][j - 1] != 1) {
                cout << "NO" << '\n';
                return;
            }
        }
    }

    cout << "YES" << '\n';
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    for (; t--;) solve();
    return 0;
}