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<vector>
#include<set>
#include<queue>
#include<map>
#include<algorithm>
#include<cstring>
#include<string>
#include<cassert>
#include<cmath>
#include<climits>
#include<iomanip>
#include<stack>
#include<unordered_map>
#include<bitset>
#include<limits>
#include<complex>
#include<array>
#include<numeric>
#include<functional>
#include<random>


using namespace std;
#define ll long long
#define ull unsigned long long
#define rep(i,m,n) for(ll (i)=(ll)(m);i<(ll)(n);i++)
#define REP(i,n) rep(i,0,n)
#define all(hoge) (hoge).begin(),(hoge).end()
#define llsize(c) ((ll)c.size())
typedef pair<ll, ll> P;
P operator+(const P a, P b) { return P(a.first + b.first, a.second + b.second); }
P operator-(const P a, P b) { return P(a.first - b.first, a.second - b.second); }


constexpr long double m_pi = 3.1415926535897932L;
constexpr ll MOD = 1000000007;
constexpr ll INF = 1LL << 61;
constexpr long double EPS = 1e-10;
template<typename T> using vector2 = vector<vector<T>>;
template<typename T> using vector3 = vector<vector2<T>>;
typedef vector<ll> Array;
typedef vector<Array> Matrix;
string operator*(const string& s, int k) {
	if (k == 0) return "";
	string p = (s + s) * (k / 2);
	if (k % 2 == 1) p += s;
	return p;
}

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; }return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; }return false; }

struct Edge {//グラフ
	int to, rev; ll cap;
	Edge(int _to, ll _cap, int _rev) {
		to = _to; cap = _cap; rev = _rev;
	}
};
typedef vector<Edge> Edges;
typedef vector<Edges> Graph;
void add_edge(Graph& G, int from, int to, ll cap, bool revFlag, ll revCap) {//最大フロー求める Ford-fulkerson
	G[from].push_back(Edge(to, cap, (ll)G[to].size() + (from == to)));
	if (revFlag)G[to].push_back(Edge(from, revCap, (ll)G[from].size() - 1));//最小カットの場合逆辺は0にする
}

ll max_flow_dfs(Graph& G, ll v, ll t, ll f, vector<bool>& used)
{
	if (v == t)
		return f;
	used[v] = true;
	for (int i = 0; i < G[v].size(); ++i) {
		Edge& e = G[v][i];
		if (!used[e.to] && e.cap > 0) {
			ll d = max_flow_dfs(G, e.to, t, min(f, e.cap), used);
			if (d > 0) {
				e.cap -= d;
				G[e.to][e.rev].cap += d;
				return d;
			}
		}
	}
	return 0;
}
//二分グラフの最大マッチングを求めたりも出来る　また二部グラフの最大独立集合は頂点数-最大マッチングのサイズ
ll max_flow(Graph& G, ll s, ll t)//O(V(V+E))
{
	ll flow = 0;
	for (;;) {
		vector<bool> used(G.size());
		REP(i, used.size())used[i] = false;
		ll f = max_flow_dfs(G, s, t, INF, used);
		if (f == 0) {
			return flow;
		}
		flow += f;
	}
}

void BellmanFord(Graph& G, ll s, Array& d, Array& negative) {//O(|E||V|)
	d.resize(G.size());
	negative.resize(G.size());
	REP(i, d.size())d[i] = INF;
	REP(i, d.size())negative[i] = false;
	d[s] = 0;
	REP(k, G.size() - 1) {
		REP(i, G.size()) {
			REP(j, G[i].size()) {
				if (d[i] != INF && d[G[i][j].to] > d[i] + G[i][j].cap) {
					d[G[i][j].to] = d[i] + G[i][j].cap;
				}
			}
		}
	}
	REP(k, G.size() - 1) {
		REP(i, G.size()) {
			REP(j, G[i].size()) {
				if (d[i] != INF && d[G[i][j].to] > d[i] + G[i][j].cap) {
					d[G[i][j].to] = d[i] + G[i][j].cap;
					negative[G[i][j].to] = true;
				}
				if (negative[i] == true)negative[G[i][j].to] = true;
			}
		}
	}
}
Array Dijkstra(Graph& g, ll s) {//O(|E|log|V|)
	Array ret(g.size(), INF);
	ret[s] = 0;
	priority_queue<P, vector<P>, greater<P>> q;
	q.push({ ret[s], s });
	while (!q.empty()) {
		auto [d, v] = q.top();
		q.pop();
		if (ret[v] < d)continue;
		for (auto e : g[v]) {
			if (chmin(ret[e.to], ret[v] + e.cap)) q.push({ ret[e.to], e.to });
		}
	}
	return ret;
}
Matrix WarshallFloyd(Graph& g) {//O(V^3)
	Matrix ret(g.size(), Array(g.size(), INF));
	REP(i, ret.size())ret[i][i] = 0;
	REP(i, g.size()) {
		for (auto e : g[i])chmin(ret[i][e.to], e.cap);
	}
	REP(i, g.size()) {
		REP(j, g.size()) {
			REP(k, g.size()) {
				if (ret[j][i] != INF && ret[i][k] != INF)chmin(ret[j][k], ret[j][i] + ret[i][k]);
			}
		}
	}
	return ret;
}
bool tsort(Graph& graph, vector<int>& order) {//トポロジカルソートO(E+V)
	int n = graph.size(), k = 0;
	vector<int> in(n);
	for (auto& es : graph)
		for (auto& e : es)in[e.to]++;
	priority_queue<int, vector<int>, greater<int>> que;
	REP(i, n)
		if (in[i] == 0)que.push(i);
	while (que.size()) {
		int v = que.top();
		que.pop();
		order.push_back(v);
		for (auto& e : graph[v])
			if (--in[e.to] == 0)que.push(e.to);
	}
	if (order.size() != n)return false;
	else return true;
}
class Lca {
public:
	const int n = 0;
	const int log2_n = 0;
	std::vector<std::vector<int>> parent;
	std::vector<int> depth;

	Lca() {}

	Lca(const Graph& g, int root)
		: n(g.size()), log2_n(log2(n) + 1), parent(log2_n, std::vector<int>(n)), depth(n) {
		dfs(g, root, -1, 0);
		for (int k = 0; k + 1 < log2_n; k++) {
			for (int v = 0; v < (int)g.size(); v++) {
				if (parent[k][v] < 0)
					parent[k + 1][v] = -1;
				else
					parent[k + 1][v] = parent[k][parent[k][v]];
			}
		}
	}

	void dfs(const Graph& g, int v, int p, int d) {
		parent[0][v] = p;
		depth[v] = d;
		for (auto& e : g[v]) {
			if (e.to != p) dfs(g, e.to, v, d + 1);
		}
	}

	int get(int u, int v) {
		if (depth[u] > depth[v]) std::swap(u, v);
		for (int k = 0; k < log2_n; k++) {
			if ((depth[v] - depth[u]) >> k & 1) {
				v = parent[k][v];
			}
		}
		if (u == v) return u;
		for (int k = log2_n - 1; k >= 0; k--) {
			if (parent[k][u] != parent[k][v]) {
				u = parent[k][u];
				v = parent[k][v];
			}
		}
		return parent[0][u];
	}
};

class UnionFind {
	vector<int> data;
	int n;
public:
	UnionFind(int size) : data(size, -1), n(size) { }
	bool merge(int x, int y) {//xとyの集合を統合する
		x = root(x); y = root(y);
		if (x != y) {
			if (data[y] < data[x]) swap(x, y);
			data[x] += data[y]; data[y] = x;
		}
		n -= (x != y);
		return x != y;
	}
	bool same(int x, int y) {//xとyが同じ集合か返す
		return root(x) == root(y);
	}
	int root(int x) {//xのルートを返す
		return data[x] < 0 ? x : data[x] = root(data[x]);
	}
	int size(int x) {//xの集合のサイズを返す
		return -data[root(x)];
	}
	int num() {//集合の数を返す
		return n;
	}
};

template<typename T, typename F>
class SegmentTree {
private:
	T identity;
	F merge;
	ll n;
	vector<T> dat;
public:
	SegmentTree(F f, T id, vector<T> v) :merge(f), identity(id) {
		int _n = v.size();
		n = 1;
		while (n < _n)n *= 2;
		dat.resize(2 * n - 1, identity);
		REP(i, _n)dat[n + i - 1] = v[i];
		for (int i = n - 2; i >= 0; i--)dat[i] = merge(dat[i * 2 + 1], dat[i * 2 + 2]);
	}
	SegmentTree(F f, T id, int _n) :merge(f), identity(id) {
		n = 1;
		while (n < _n)n *= 2;
		dat.resize(2 * n - 1, identity);
	}
	void set_val(int i, T x) {
		i += n - 1;
		dat[i] = x;
		while (i > 0) {
			i = (i - 1) / 2;
			dat[i] = merge(dat[i * 2 + 1], dat[i * 2 + 2]);
		}
	}
	T query(int l, int r) {
		T left = identity, right = identity;
		l += n - 1; r += n - 1;
		while (l < r) {
			if ((l & 1) == 0)left = merge(left, dat[l]);
			if ((r & 1) == 0)right = merge(dat[r - 1], right);
			l = l / 2;
			r = (r - 1) / 2;
		}
		return merge(left, right);
	}
};


template< typename T >
class FenwickTree {
	vector< T > data;
	int n;
	int p;
public:
	FenwickTree(int n) :n(n) {
		data.resize(n + 1LL, 0);
		p = 1;
		while (p < data.size())p *= 2;
	}
	T sum(int k) {
		T ret = 0;
		for (; k > 0; k -= k & -k) ret += data[k];
		return (ret);
	}

	T sum(int a, int b) { return sum(b) - sum(a); }//[a,b)

	void add(int k, T x) {
		for (++k; k <= n; k += k & -k) data[k] += x;
	}

	int lower_bound(ll w) {
		if (w <= 0)return -1;
		int x = 0;
		for (int k = p / 2; k > 0; k /= 2) {
			if (x + k <= n && data[x + k] < w)w -= data[x + k], x += k;
		}
		return x;
	}
};



//約数求める //約数
template<typename T>
vector<T> divisor(T n) {
	vector<T> ret;
	for (T i = 1; i * i <= n; i++) {
		if (n % i == 0) {
			ret.push_back(i);
			if (i * i != n) ret.push_back(n / i);
		}
	}
	sort(ret.begin(), ret.end());
	return ret;
}

template<typename T>
vector<pair<T, int>> prime_factorization(T n) {
	vector<pair<T, int>> ret;
	for (T i = 2; i * i <= n; i++) {
		if (n % i == 0) {
			ret.push_back({ i,0 });
			while (n % i == 0) {
				n /= i;
				ret[ret.size() - 1].second++;
			}
		}
	}
	if (n != 1)ret.push_back({ n,1 });
	return ret;
}

inline ll mod_pow(ll x, ll n, ll mod) {
	ll res = 1;
	while (n > 0) {
		if (n & 1) res = res * x % mod;
		x = x * x % mod;
		n >>= 1;
	}
	return res;
}

inline ll mod_inv(ll x, ll mod) {
	return mod_pow(x, mod - 2, mod);
}

class Combination {
public:
	Array fact;
	Array fact_inv;
	ll mod;
	//if n >= mod use lucas 
	ll nCr(ll n, ll r) {
		if (n < r)return 0;
		if (n < mod)return ((fact[n] * fact_inv[r] % mod) * fact_inv[n - r]) % mod;

		ll ret = 1;
		while (n || r) {
			ll _n = n % mod, _r = r % mod;
			n /= mod; r /= mod;
			(ret *= nCr(_n, _r)) %= mod;
		}
		return ret;
	}
	ll nPr(ll n, ll r) {
		return (fact[n] * fact_inv[n - r]) % mod;
	}
	ll nHr(ll n, ll r) {
		return nCr(r + n - 1, r);
	}
	Combination(ll _n, ll _mod) {
		mod = _mod;
		ll n = min(_n + 1, mod);
		fact.resize(n);
		fact[0] = 1;
		REP(i, n - 1) {
			fact[i + 1] = (fact[i] * (i + 1LL)) % mod;
		}
		fact_inv.resize(n);
		fact_inv[n - 1] = mod_inv(fact[n - 1], mod);
		for (int i = n - 1; i > 0; i--) {
			fact_inv[i - 1] = fact_inv[i] * i % mod;
		}
	}
};

ll popcount(ll x) {
	x = (x & 0x5555555555555555) + (x >> 1 & 0x5555555555555555);
	x = (x & 0x3333333333333333) + (x >> 2 & 0x3333333333333333);
	x = (x & 0x0F0F0F0F0F0F0F0F) + (x >> 4 & 0x0F0F0F0F0F0F0F0F);
	x = (x & 0x00FF00FF00FF00FF) + (x >> 8 & 0x00FF00FF00FF00FF);
	x = (x & 0x0000FFFF0000FFFF) + (x >> 16 & 0x0000FFFF0000FFFF);
	x = (x & 0x00000000FFFFFFFF) + (x >> 32 & 0x00000000FFFFFFFF);

	return x;
}



constexpr ll mod = 998244353;





int main() {
	ios::sync_with_stdio(false);
	std::cin.tie(0);
	std::cout.tie(0);


	
	ll n, k;
	cin >> n >> k;
	Array p(n), rev(n + 1, -1);
	REP(i, n)cin >> p[i], rev[p[i]] = i;
	ll cnt = 1;
	REP(i, n - 1) {
		if (rev[p[i] + 1] > rev[p[i + 1] + 1])cnt++;
	}
	if (cnt > k)cout << 0 << "\n";
	else {
		Combination comb(404040, mod);
		cout << comb.nCr(k - cnt + n, n) << "\n";
	}






	return 0;
}