Problem: Ashish has two strings a and b, each of length n, and an integer k. The strings only contain lowercase English letters.

He wants to convert string a into string b by performing some (possibly zero) operations on a.

In one move, he can either 

  * choose an index i (1 ≤ i≤ n-1) and swap a_i and a_{i+1}, or 
  * choose an index i (1 ≤ i ≤ n-k+1) and if a_i, a_{i+1}, …, a_{i+k-1} are all equal to some character c (c ≠ 'z'), replace each one with the next character (c+1), that is, 'a' is replaced by 'b', 'b' is replaced by 'c' and so on. 



Note that he can perform any number of operations, and the operations can only be performed on string a. 

Help Ashish determine if it is possible to convert string a into b after performing some (possibly zero) operations on it.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The description of each test case is as follows.

The first line of each test case contains two integers n (2 ≤ n ≤ 10^6) and k (1 ≤ k ≤ n).

The second line of each test case contains the string a of length n consisting of lowercase English letters.

The third line of each test case contains the string b of length n consisting of lowercase English letters.

It is guaranteed that the sum of values n among all test cases does not exceed 10^6.

Output

For each test case, print "Yes" if Ashish can convert a into b after some moves, else print "No".

You may print the letters of the answer in any case (upper or lower).

Example

Input


4
3 3
abc
bcd
4 2
abba
azza
2 1
zz
aa
6 2
aaabba
ddddcc


Output


No
Yes
No
Yes

Note

In the first test case it can be shown that it is impossible to convert a into b.

In the second test case,

"abba" \xrightarrow{inc} "acca" \xrightarrow{inc} … \xrightarrow{inc} "azza".

Here "swap" denotes an operation of the first type, and "inc" denotes an operation of the second type.

In the fourth test case,

"aaabba" \xrightarrow{swap} "aaabab" \xrightarrow{swap} "aaaabb" \xrightarrow{inc} … \xrightarrow{inc} "ddaabb" \xrightarrow{inc} … \xrightarrow{inc} "ddddbb" \xrightarrow{inc} … \xrightarrow{inc} "ddddcc".

Solution (in CPP):

Solution:
// Problem: C. String Equality
// Contest: Codeforces - Codeforces Round #685 (Div. 2)
// URL: https://codeforces.com/group/zglzo6fOCZ/contest/1451/problem/C
// Memory Limit: 256 MB
// Time Limit: 2000 ms

#include <assert.h>

#include <algorithm>
#include <array>
#include <atomic>
#include <bitset>
#include <chrono>
#include <complex>
#include <condition_variable>
#include <deque>
#include <exception>
#include <forward_list>
#include <fstream>
#include <functional>
#include <future>
#include <initializer_list>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <mutex>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <random>
#include <ratio>
#include <regex>
#include <scoped_allocator>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <system_error>
#include <thread>
#include <tuple>
#include <type_traits>
#include <typeindex>
#include <typeinfo>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <valarray>
#include <vector>

using namespace std;

#define LOCAL
#define time cout << (0.1 * clock()) / CLOCKS_PER_SEC << endl;
#define countBits(x) __builtin_popcount(ll(x))
#define countZeroesAtBegin(x) __builtin_clz(ll(x))
#define countZeroesAtEnd(x) __builtin_ctz(ll(x))
#define last(x) x[x.end() - x.begin() - 1]
#define pb push_back
#define bg begin
#define ff first
#define ss second
#define pi 3.1415926535897932384626
#define infll 0x3f3f3f3f3f3f3f3f
using ll = long long;
using pl = pair<ll, ll>;
using vll = vector<ll>;
using vpl = vector<pl>;
using mat = vector<vll>;
const ll MAX = 1e5;
const ll mod = 1e9 + 7;
#define all(x) x.begin(), x.end()
#define clr(x) memset(x, 0, sizeof(x))
#define sortall(x) sort(all(x))

// For Loop templates
#define F_OR(i, a, b, s) \
    for (ll i = (a); (s) > 0 ? i < (b) : i >= (b); i += (s))
#define F_OR1(e) F_OR(i, 0, e, 1)
#define F_OR2(i, e) F_OR(i, 0, e, 1)
#define F_OR3(i, b, e) F_OR(i, b, e, 1)
#define F_OR4(i, b, e, s) F_OR(i, b, e, s)
#define GET5(a, b, c, d, e, ...) e
#define F_ORC(...) GET5(__VA_ARGS__, F_OR4, F_OR3, F_OR2, F_OR1)
#define FOR(...)       \
    F_ORC(__VA_ARGS__) \
    (__VA_ARGS__)
#define EACH(x, a) for (auto &x : a)
#define itr(it, a) for (auto it = a.begin(); it != a.end(); it++)

template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
    EACH(ele, v) { cout << ele << " "; }
    return os;
}

template <typename T, typename U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
    return os << '(' << p.ff << "," << p.ss << ')';
}

template <typename T>
istream &operator>>(istream &in, vector<T> &v) {
    EACH(ele, v) { cin >> ele; }
    return in;
}

// Debugging Template
template <typename A, typename B>
string to_string(pair<A, B> p);

template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p);

template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p);

string to_string(const string &s) { return '"' + s + '"'; }

string to_string(const char *s) { return to_string((string)s); }

string to_string(bool b) { return (b ? "true" : "false"); }

string to_string(vector<bool> v) {
    bool first = true;
    string res = "{";
    for (int i = 0; i < static_cast<int>(v.size()); i++) {
        if (!first) {
            res += ", ";
        }
        first = false;
        res += to_string(v[i]);
    }
    res += "}";
    return res;
}

template <size_t N>
string to_string(bitset<N> v) {
    string res = "";
    for (size_t i = 0; i < N; i++) {
        res += static_cast<char>('0' + v[i]);
    }
    return res;
}

template <typename A>
string to_string(A v) {
    bool first = true;
    string res = "{";
    for (const auto &x : v) {
        if (!first) {
            res += ", ";
        }
        first = false;
        res += to_string(x);
    }
    res += "}";
    return res;
}

template <typename A, typename B>
string to_string(pair<A, B> p) {
    return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}

template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p) {
    return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
           to_string(get<2>(p)) + ")";
}

template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p) {
    return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
           to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")";
}

void debug_out() { cerr << endl; }

template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
    cerr << " " << to_string(H);
    debug_out(T...);
}

#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif

void solve() {
    ll N, K;
    cin >> N >> K;
    string A, B;
    cin >> A >> B;
    array<int, 27> have{}, need{};
    EACH(it, A) { have[it - 'a'] += 1; }
    EACH(it, B) { need[it - 'a'] += 1; }
    bool flag = false;
    FOR(26) {
        if (have[i] < need[i] || (have[i] -= need[i]) % K) {
            flag = true;
        }
        have[i + 1] += have[i];
    }
    cout << (!flag ? "Yes\n" : "No\n");
}

int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ll tc = 1;
    cin >> tc;
    while (tc--) {
        solve();
    }
    return 0;
}