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.
#include <iostream>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstdlib>
#include <cassert>
#include <vector>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <string>
#include <algorithm>
#include <utility>
#include <complex>
#define rep(x, s, t) for(llint (x) = (s); (x) <= (t); (x)++)
#define per(x, s, t) for(llint (x) = (s); (x) >= (t); (x)--)
#define reps(x, s) for(llint (x) = 0; (x) < (llint)(s).size(); (x)++)
#define chmin(x, y) (x) = min((x), (y))
#define chmax(x, y) (x) = max((x), (y))
#define sz(x) ((ll)(x).size())
#define ceil(x, y) (((x)+(y)-1) / (y))
#define all(x) (x).begin(),(x).end()
#define outl(...) dump_func(__VA_ARGS__)
#define inf 1e18

using namespace std;

typedef long long llint;
typedef long long ll;
typedef pair<ll, ll> P;

struct edge{
	ll to, cost;
	edge(){}
	edge(ll a, ll b){ to = a, cost = b;}
};
const ll dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};

//const ll mod = 1000000007;
const ll mod = 998244353;

struct mint{
	ll x = 0;
	mint(ll y = 0){x = y; if(x < 0 || x >= mod) x = (x%mod+mod)%mod;}
	mint(const mint &ope) {x = ope.x;}
	
	mint operator-(){return mint(-x);}
	mint operator+(const mint &ope){return mint(x) += ope;}
	mint operator-(const mint &ope){return mint(x) -= ope;}
	mint operator*(const mint &ope){return mint(x) *= ope;}
	mint operator/(const mint &ope){return mint(x) /= ope;}
	mint& operator+=(const mint &ope){
		x += ope.x;
		if(x >= mod) x -= mod;
		return *this;
	}
	mint& operator-=(const mint &ope){
		x += mod - ope.x;
		if(x >= mod) x -= mod;
		return *this;
	}
	mint& operator*=(const mint &ope){
		x *= ope.x, x %= mod;
		return *this;
	}
	mint& operator/=(const mint &ope){
		ll n = mod-2; mint mul = ope;
		while(n){
			if(n & 1) *this *= mul;
			mul *= mul;
			n >>= 1;
		}
		return *this;
	}
	mint inverse(){return mint(1) / *this;}
	bool operator ==(const mint &ope){return x == ope.x;}
	bool operator !=(const mint &ope){return x != ope.x;}
};
mint modpow(mint a, ll n){
	if(n == 0) return mint(1);
	if(n % 2) return a * modpow(a, n-1);
	else return modpow(a*a, n/2);
}
istream& operator >>(istream &is, mint &ope){
	ll t; is >> t, ope.x = t;
	return is;
}
ostream& operator <<(ostream &os, mint &ope){return os << ope.x;}
ostream& operator <<(ostream &os, const mint &ope){return os << ope.x;}

bool exceed(ll x, ll y, ll m){return x >= m / y + 1;}
void mark(){ cout << "*" << endl; }
void yes(){ cout << "YES" << endl; }
void no(){ cout << "NO" << endl; }
ll sgn(ll x){ if(x > 0) return 1; if(x < 0) return -1; return 0;}
ll gcd(ll a, ll b){if(b == 0) return a; return gcd(b, a%b);}
ll lcm(ll a, ll b){return a/gcd(a, b)*b;}
ll digitnum(ll x, ll b = 10){ll ret = 0; for(; x; x /= b) ret++; return ret;}
ll digitsum(ll x, ll b = 10){ll ret = 0; for(; x; x /= b) ret += x % b; return ret;}
string lltos(ll x){string ret; for(;x;x/=10) ret += x % 10 + '0'; reverse(ret.begin(), ret.end()); return ret;}
ll stoll(string &s){ll ret = 0; for(auto c : s) ret *= 10, ret += c - '0'; return ret;}
template<typename T>
void uniq(T &vec){ sort(vec.begin(), vec.end()); vec.erase(unique(vec.begin(), vec.end()), vec.end());}

template<typename T>
ostream& operator << (ostream& os, vector<T>& vec) {
	for(int i = 0; i < vec.size(); i++) os << vec[i] << (i + 1 == vec.size() ? "" : " ");
	return os;
}
template<typename T>
ostream& operator << (ostream& os, deque<T>& deq) {
	for(int i = 0; i < deq.size(); i++) os << deq[i] << (i + 1 == deq.size() ? "" : " ");
	return os;
}
template<typename T, typename U>
ostream& operator << (ostream& os, pair<T, U>& pair_var) {
	os << "(" << pair_var.first << ", " << pair_var.second << ")";
	return os;
}
template<typename T, typename U>
ostream& operator << (ostream& os, const pair<T, U>& pair_var) {
	os << "(" << pair_var.first << ", " << pair_var.second << ")";
	return os;
}
template<typename T, typename U>
ostream& operator << (ostream& os, map<T, U>& map_var) {
	for(typename map<T, U>::iterator itr = map_var.begin(); itr != map_var.end(); itr++) {
		os << "(" << itr->first << ", " << itr->second << ")";
		itr++; if(itr != map_var.end()) os << ","; itr--;
	}
	return os;
}
template<typename T>
ostream& operator << (ostream& os, set<T>& set_var) {
	for(typename set<T>::iterator itr = set_var.begin(); itr != set_var.end(); itr++) {
		os << *itr; ++itr; if(itr != set_var.end()) os << " "; itr--;
	}
	return os;
}
template<typename T>
ostream& operator << (ostream& os, multiset<T>& set_var) {
	for(typename multiset<T>::iterator itr = set_var.begin(); itr != set_var.end(); itr++) {
		os << *itr; ++itr; if(itr != set_var.end()) os << " "; itr--;
	}
	return os;
}
template<typename T>
void outa(T a[], ll s, ll t){
	for(ll i = s; i <= t; i++){ cout << a[i]; if(i < t) cout << " ";}
	cout << endl;
}
void dump_func() {cout << endl;}
template <class Head, class... Tail>
void dump_func(Head &&head, Tail &&... tail) {
	cout << head;
	if(sizeof...(Tail) > 0) cout << " ";
	dump_func(std::move(tail)...);
}


const int FACT_MAX = 600005;
mint fact[FACT_MAX], fact_inv[FACT_MAX];

void make_fact()
{
	mint val = 1;
	fact[0] = 1;
	for(int i = 1; i < FACT_MAX; i++){
		val *= i;
		fact[i] = val;
	}
	fact_inv[FACT_MAX-1] = fact[FACT_MAX-1].inverse();
	for(int i = FACT_MAX-2; i >= 0; i--){
		fact_inv[i] = fact_inv[i+1] * (i+1);
	}
}

mint comb(ll n, ll k)
{
	mint ret = 1;
	ret *= fact[n];
	ret *= fact_inv[k];
	ret *= fact_inv[n-k];
	return ret;
}


ll n, k;
ll p[200005], q[200005];

int main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	make_fact();
	
	cin >> n >> k;
	rep(i, 1, n) cin >> p[i], p[i]++;
	rep(i, 1, n) q[p[i]] = i;
	
	k--;
	rep(i, 1, n-1){
		if(q[p[i]+1] > q[p[i+1]+1]) k--;
	}
	if(k < 0) outl(0);
	else outl(comb(n+k, k));
	
	return 0;
}