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 <sstream>
#include <iomanip>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <bitset>
#include <string>
#include <cstring>
#include <cmath>
#include <cstdio>
#define all(v) (v).begin(), (v).end()
#define unq(v) (v).erase(unique(all(v)), (v).end())
#define vec vector
#define pii pair <int, int>
#define pli pair <ll, int>
#define pil pair <int, ll>
#define pll2 pair <ll, ll>
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define MULTI int T; cin >> T; while(T--)
#define sqr(x) ((x) * (x))
#define yesno(x) cout << ((x) ? "Yes" : "No") << endl
#define test cerr << '!' << endl;
using namespace std;

typedef long long ll;
typedef unsigned long long ull;

void main0 ();
int main () {
	clock_t start, end;
	#ifndef ONLINE_JUDGE
		freopen("C:\\Users\\98497\\Desktop\\code\\file.in", "r", stdin);
		start = clock();
	#endif
	ios::sync_with_stdio(false); cin.tie(0);
	main0();
	#ifndef ONLINE_JUDGE
		end = clock();
		cout << "( " << (end - start) << " ms )" << endl;
		fclose(stdin);
	#endif
	return 0;
}

const int dx[8] = {0, 1, -1, 0, 1, 1, -1, -1};
const int dy[8] = {1, 0, 0, -1, 1, -1, -1, 1};
const int N = 2e5 + 5;
const int M = 1e5;
const int INF = 0x3f3f3f3f;
const ll INFF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-6;
const double Pi = acos(-1.0);

ll qpow (ll n, ll m) {
	ll ret = 1;
	while (m) {
		if (m & 1) ret = ret * n % mod;
		n = n * n % mod;
		m >>= 1;
	}
	return ret;
}
ll getinv (ll a) {
	return qpow(a, mod - 2);
}

ll fac[N] = {1}, inv[N];
void init () {
	for (int i = 1;i < N;++i) {
		fac[i] = fac[i - 1] * i % mod;
	}
	inv[N - 1] = getinv(fac[N - 1]);
	for (int i = N - 2;i >= 0;--i) {
		inv[i] = inv[i + 1] * (i + 1) % mod;
	}
}

ll cmn (ll m, ll n) {
	if (m < n) return 0;
	ll ret = fac[m];
	ret = ret * inv[n] % mod;
	ret = ret * inv[m - n] % mod;
	return ret;
}

ll n, k;
int sa[N];
int rnk[N];

void main0 () {
	init();
	
	cin >> n >> k;
	for (int i = 1;i <= n;++i) {
		cin >> sa[i];
		rnk[sa[i] + 1] = i;
	}
	rnk[n + 1] = -INF;
	
	int block = 1;
	for (int i = 1;i <= n - 1;++i) {
		int u = sa[i] + 1;
		int v = sa[i + 1] + 1;
		if (rnk[u + 1] > rnk[v + 1]) block++;
	}
	//cerr << block << endl;
	ll ans = 0;
	for (int i = block;i <= n;++i) {
		//cout << k << ' ' << i << ' ' << n - block << ' ' << n - i << endl;
		ans += cmn(k, i) * cmn(n - block, n - i);
		ans %= mod;
	}
	cout << ans << endl;
}