Once upon a time, Oolimry saw a suffix array. He wondered how many strings can produce this suffix array. 

More formally, given a suffix array of length n and having an alphabet size k, count the number of strings that produce such a suffix array. 

Let s be a string of length n. Then the i-th suffix of s is the substring s[i … n-1]. A suffix array is the array of integers that represent the starting indexes of all the suffixes of a given string, after the suffixes are sorted in the lexicographic order. For example, the suffix array of oolimry is [3,2,4,1,0,5,6] as the array of sorted suffixes is [imry,limry,mry,olimry,oolimry,ry,y]. 

A string x is lexicographically smaller than string y, if either x is a prefix of y (and x≠ y), or there exists such i that x_i < y_i, and for any 1≤ j < i , x_j = y_j.

Input

The first line contain 2 integers n and k (1 ≤ n ≤ 200000,1 ≤ k ≤ 200000) — the length of the suffix array and the alphabet size respectively.

The second line contains n integers s_0, s_1, s_2, …, s_{n-1} (0 ≤ s_i ≤ n-1) where s_i is the i-th element of the suffix array i.e. the starting position of the i-th lexicographically smallest suffix. It is guaranteed that for all 0 ≤ i< j ≤ n-1, s_i ≠ s_j.

Output

Print how many strings produce such a suffix array. Since the number can be very large, print the answer modulo 998244353.

Examples

Input


3 2
0 2 1


Output


1

Input


5 1
0 1 2 3 4


Output


0

Input


6 200000
0 1 2 3 4 5


Output


822243495

Input


7 6
3 2 4 1 0 5 6


Output


36

Note

In the first test case, "abb" is the only possible solution. 

In the second test case, it can be easily shown no possible strings exist as all the letters have to be equal. 

In the fourth test case, one possible string is "ddbacef".

Please remember to print your answers modulo 998244353.
//NeeD FoR AC!
#include <bits/stdc++.h>

/*
#pragma GCC optimize("Ofast")
#pragma GCC optimize ("unroll-loops")
#pragma GCC optimize("O0")
#pragma GCC optimize("O1")
#pragma GCC optimize("Ofast","unroll-loops","O3")
*/

using namespace std ;
 
typedef long long ll ;
typedef long double ld ;
typedef pair<ld ,int> pfi ;
typedef pair<int ,int> pii ; 
typedef pair<ll ,pii> pip ;

#define fast_io                ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define debug(x)               cerr << #x << " : " << x << endl
#define debugP(x ,y)           cerr << #x << "," << #y << " : " << x << "," << y << endl
#define here                   cerr << "here\n"
#define MP                     make_pair
#define F                      first
#define S                      second
 
constexpr int MAXN = 5e5 + 2 ;
constexpr int MOD = 998244353 ;
constexpr int INF = 1e9 + 9 ;
constexpr int LOG = 2e1 + 2 ;

int fact[MAXN] ,invfact[MAXN] ,n ,a[MAXN] ,k ,ans;
map <int ,int> idx ; 
 
int add(int a ,int b){
	int c = a + b ; 
	if (c >= MOD){
		c -= MOD ; 
	}
	if (c < 0){
		c += MOD ; 
	}
	return c ; 
}

int mul(int a ,int b){
	return 1ll * a * b % MOD ; 
}

int pwr(int a ,int b){
	if (!b){
		return 1 ; 
	}
	int tmp = pwr(a ,b / 2) ; 
	tmp = mul(tmp ,tmp) ; 
	if (b & 1){
		tmp = mul(tmp ,a) ; 
	}
	return tmp ; 
}
int C(int n ,int k){
	if (k > n || k < 0){
		return 0 ; 
	}
	int tmp = fact[n] ; 
	tmp = mul(tmp ,invfact[k]) ; 
	tmp = mul(tmp ,invfact[n - k]) ; 
	return tmp ; 
}

int inv(int p){
	return pwr(p ,MOD - 2) ; 
}

int main (){
	fast_io ;
	cin >> n >> k ; 
	int cnt = 1 ; 
	for (int i = 1; i <= n; i++){
		cin >> a[i] ; 
		idx[a[i]] = i ; 
	}
	for (int i = 2; i <= n; i++){
		if (idx[a[i] + 1] < idx[a[i - 1] + 1]){ 
			cnt++ ; 
		}
	}
	fact[0] = 1; 
	invfact[0] = inv(1) ; 
	for (int i = 1; i < MAXN - 1; i++){
		fact[i] = mul(fact[i - 1] ,i) ; 
		invfact[i] = inv(fact[i]) ; 
	} 
	for (int i = 0; i <= k - cnt; i++){ 
		ans = add(C(n + i - 1 ,n - 1) ,ans) ; 
	}
	cout << ans ; 
	return 0 ;
}


