Problem: The only difference between the two versions of the problem is that there are no updates in the easy version.

There are n spools of thread placed on the rim of a circular table. The spools come in two types of thread: the first thread is black and the second thread is white.

For any two spools of the same color, you can attach them with a thread of that color in a straight line segment. Define a matching as a way to attach spools together so that each spool is attached to exactly one other spool.

Coloring is an assignment of colors (white and black) to the spools. A coloring is called valid if it has at least one matching. That is if the number of black spools and the number of white spools are both even.

Given a matching, we can find the number of times some white thread intersects some black thread. We compute the number of pairs of differently colored threads that intersect instead of the number of intersection points, so one intersection point may be counted multiple times if different pairs of threads intersect at the same point. If c is a valid coloring, let f(c) denote the minimum number of such intersections out of all possible matchings.

<image> The circle above is described by the coloring bwbbbwww. After matching the spools as shown, there is one intersection between differently colored threads. It can be proven that it is the minimum possible, so f(bwbbbwww) = 1. 

You are given a string s representing an unfinished coloring, with black, white, and uncolored spools. A coloring c is called s-reachable if you can achieve it by assigning colors to the uncolored spools of s without changing the others.

A coloring c is chosen uniformly at random among all valid, s-reachable colorings. Compute the [expected value](https://en.wikipedia.org/wiki/Expected_value) of f(c). You should find it by modulo 998244353.

There will be m updates to change one character of s. After each update, you should again compute the expected value of f(c).

We can show that each answer can be written in the form p/q where p and q are relatively prime integers and qnot≡ 0\pmod{998244353}. The answer by modulo 998244353 is equal to (p⋅ q^{-1}) modulo 998244353.

Input

The first line contains two integers n, m (2≤ n≤ 2⋅ 10^5, n is even, 0≤ m≤ 2⋅ 10^5) — the number of spools and the number of updates, respectively.

The second line contains a string s of length n — the unfinished coloring of the spools. The i-th character will be 'w', 'b', or '?', describing if the i-th spool is white, black, or uncolored, respectively.

Each of the next m lines contains an integer i (1 ≤ i ≤ n) — the position of the character in s to be updated, and a character c (c ∈ \{w, b, ?\}) — the new color of the spool i after the update.

It is guaranteed there exists at least one uncolored spool initially and after each update.

Output

Print m+1 lines: the expected value of f(c) initially and after each update. All values should be found by modulo 998244353.

Examples

Input


8 0
bwbb?www


Output


1


Input


10 3
???ww?wb??
4 ?
5 ?
2 w


Output


436731905
218365953
374341633
530317313


Input


4 3
bw?b
1 w
2 b
1 w


Output


0
0
1
1

Note

The first test corresponds closely to the image. Coloring '?' as 'w' does not create a valid coloring because the number of black spools is odd. Then the only reachable valid coloring is 'bwbbbwww' and f(bwbbbwww) = 1, so the expected value is 1.

In the second test, the string after each update is:

  1. ????w?wb??
  2. ??????wb??
  3. ?w????wb??



In the third test, the string after each update is:

  1. ww?b
  2. wb?b
  3. wb?b

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
 
typedef long long ll;
typedef long double ld;
typedef double db; 
typedef string str; 

typedef pair<int,int> pi;
typedef pair<ll,ll> pl; 
typedef pair<db,db> pd; 

typedef vector<int> vi; 
typedef vector<bool> vb; 
typedef vector<ll> vl; 
typedef vector<db> vd; 
typedef vector<str> vs; 
typedef vector<pi> vpi;
typedef vector<pl> vpl; 
typedef vector<pd> vpd; 

#define mp make_pair
#define f first
#define s second
#define sz(x) (int)(x).size()
#define all(x) begin(x), end(x)
#define rall(x) (x).rbegin(), (x).rend() 
#define sor(x) sort(all(x)) 
#define rsz resize
#define ins insert 
#define ft front() 
#define bk back()
#define pf push_front 
#define pb push_back
#define eb emplace_back 
#define lb lower_bound 
#define ub upper_bound 

#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)

const int MOD = 998244353;
const int MX = 2e5+5; 
const ll INF = 1e18; 
const ld PI = acos((ld)-1);
const int xd[4] = {1,0,-1,0}, yd[4] = {0,1,0,-1}; 
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count()); 

template<class T> bool ckmin(T& a, const T& b) { 
	return b < a ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) { 
	return a < b ? a = b, 1 : 0; } 
constexpr int pct(int x) { return __builtin_popcount(x); } 
constexpr int bits(int x) { return 31-__builtin_clz(x); } // floor(log2(x)) 
ll cdiv(ll a, ll b) { return a/b+((a^b)>0&&a%b); } // divide a by b rounded up
ll fdiv(ll a, ll b) { return a/b-((a^b)<0&&a%b); } // divide a by b rounded down
ll half(ll x) { return fdiv(x,2); }

template<class T, class U> T fstTrue(T lo, T hi, U f) { 
	// note: if (lo+hi)/2 is used instead of half(lo+hi) then this will loop infinitely when lo=hi
	hi ++; assert(lo <= hi); // assuming f is increasing
	while (lo < hi) { // find first index such that f is true 
		T mid = half(lo+hi);
		f(mid) ? hi = mid : lo = mid+1; 
	} 
	return lo;
}
template<class T, class U> T lstTrue(T lo, T hi, U f) {
	lo --; assert(lo <= hi); // assuming f is decreasing
	while (lo < hi) { // find first index such that f is true 
		T mid = half(lo+hi+1);
		f(mid) ? lo = mid : hi = mid-1;
	} 
	return lo;
}
template<class T> void remDup(vector<T>& v) { 
	sort(all(v)); v.erase(unique(all(v)),end(v)); }

// INPUT
template<class A> void re(complex<A>& c);
template<class A, class B> void re(pair<A,B>& p);
template<class A> void re(vector<A>& v);
template<class A, size_t SZ> void re(array<A,SZ>& a);

template<class T> void re(T& x) { cin >> x; }
void re(db& d) { str t; re(t); d = stod(t); }
void re(ld& d) { str t; re(t); d = stold(t); }
template<class H, class... T> void re(H& h, T&... t) { re(h); re(t...); }

template<class A> void re(complex<A>& c) { A a,b; re(a,b); c = {a,b}; }
template<class A, class B> void re(pair<A,B>& p) { re(p.f,p.s); }
template<class A> void re(vector<A>& x) { trav(a,x) re(a); }
template<class A, size_t SZ> void re(array<A,SZ>& x) { trav(a,x) re(a); }

// TO_STRING
#define ts to_string
str ts(char c) { return str(1,c); }
str ts(const char* s) { return (str)s; }
str ts(str s) { return s; }
str ts(bool b) { 
	#ifdef LOCAL
		return b ? "true" : "false"; 
	#else 
		return ts((int)b);
	#endif
}
template<class A> str ts(complex<A> c) { 
	stringstream ss; ss << c; return ss.str(); }
str ts(vector<bool> v) {
	str res = "{"; F0R(i,sz(v)) res += char('0'+v[i]);
	res += "}"; return res; }
template<size_t SZ> str ts(bitset<SZ> b) {
	str res = ""; F0R(i,SZ) res += char('0'+b[i]);
	return res; }
template<class A, class B> str ts(pair<A,B> p);
template<class T> str ts(T v) { // containers with begin(), end()
	#ifdef LOCAL
		bool fst = 1; str res = "{";
		for (const auto& x: v) {
			if (!fst) res += ", ";
			fst = 0; res += ts(x);
		}
		res += "}"; return res;
	#else
		bool fst = 1; str res = "";
		for (const auto& x: v) {
			if (!fst) res += " ";
			fst = 0; res += ts(x);
		}
		return res;

	#endif
}
template<class A, class B> str ts(pair<A,B> p) {
	#ifdef LOCAL
		return "("+ts(p.f)+", "+ts(p.s)+")"; 
	#else
		return ts(p.f)+" "+ts(p.s);
	#endif
}

// OUTPUT
template<class A> void pr(A x) { cout << ts(x); }
template<class H, class... T> void pr(const H& h, const T&... t) { 
	pr(h); pr(t...); }
void ps() { pr("\n"); } // print w/ spaces
template<class H, class... T> void ps(const H& h, const T&... t) { 
	pr(h); if (sizeof...(t)) pr(" "); ps(t...); }

// DEBUG
void DBG() { cerr << "]" << endl; }
template<class H, class... T> void DBG(H h, T... t) {
	cerr << ts(h); if (sizeof...(t)) cerr << ", ";
	DBG(t...); }
#ifdef LOCAL // compile with -DLOCAL, chk -> fake assert
	#define dbg(...) cerr << "Line(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)
	#define chk(...) if (!(__VA_ARGS__)) cerr << "Line(" << __LINE__ << ") -> function(" \
		 << __FUNCTION__  << ") -> CHK FAILED: (" << #__VA_ARGS__ << ")" << "\n", exit(0);
#else
	#define dbg(...) 0
	#define chk(...) 0
#endif

// FILE I/O
void setIn(str s) { freopen(s.c_str(),"r",stdin); }
void setOut(str s) { freopen(s.c_str(),"w",stdout); }
void unsyncIO() { cin.tie(0)->sync_with_stdio(0); }
void setIO(str s = "") {
	unsyncIO();
	// cin.exceptions(cin.failbit); 
	// throws exception when do smth illegal
	// ex. try to read letter into int
	if (sz(s)) { setIn(s+".in"), setOut(s+".out"); } // for USACO
}

/**
 * Description: modular arithmetic operations 
 * Source: 
	* KACTL
	* https://codeforces.com/blog/entry/63903
	* https://codeforces.com/contest/1261/submission/65632855 (tourist)
	* https://codeforces.com/contest/1264/submission/66344993 (ksun)
	* also see https://github.com/ecnerwala/cp-book/blob/master/src/modnum.hpp (ecnerwal)
 * Verification: 
	* https://open.kattis.com/problems/modulararithmetic
 */

template<int MOD, int RT> struct mint {
	static const int mod = MOD;
	static constexpr mint rt() { return RT; } // primitive root for FFT
	int v; explicit operator int() const { return v; } // explicit -> don't silently convert to int
	mint() { v = 0; }
	mint(ll _v) { v = (-MOD < _v && _v < MOD) ? _v : _v % MOD;
		if (v < 0) v += MOD; }
	friend bool operator==(const mint& a, const mint& b) { 
		return a.v == b.v; }
	friend bool operator!=(const mint& a, const mint& b) { 
		return !(a == b); }
	friend bool operator<(const mint& a, const mint& b) { 
		return a.v < b.v; }
	friend void re(mint& a) { ll x; re(x); a = mint(x); }
	friend str ts(mint a) { return ts(a.v); }
   
	mint& operator+=(const mint& m) { 
		if ((v += m.v) >= MOD) v -= MOD; 
		return *this; }
	mint& operator-=(const mint& m) { 
		if ((v -= m.v) < 0) v += MOD; 
		return *this; }
	mint& operator*=(const mint& m) { 
		v = (ll)v*m.v%MOD; return *this; }
	mint& operator/=(const mint& m) { return (*this) *= inv(m); }
	friend mint pow(mint a, ll p) {
		mint ans = 1; assert(p >= 0);
		for (; p; p /= 2, a *= a) if (p&1) ans *= a;
		return ans; }
	friend mint inv(const mint& a) { assert(a.v != 0); 
		return pow(a,MOD-2); }
		
	mint operator-() const { return mint(-v); }
	mint& operator++() { return *this += 1; }
	mint& operator--() { return *this -= 1; }
	friend mint operator+(mint a, const mint& b) { return a += b; }
	friend mint operator-(mint a, const mint& b) { return a -= b; }
	friend mint operator*(mint a, const mint& b) { return a *= b; }
	friend mint operator/(mint a, const mint& b) { return a /= b; }
};

typedef mint<MOD,5> mi; // 5 is primitive root for both common mods
typedef vector<mi> vmi;
typedef pair<mi,mi> pmi;
typedef vector<pmi> vpmi;

vector<vmi> scmb; // small combinations
void genComb(int SZ) {
	scmb.assign(SZ,vmi(SZ)); scmb[0][0] = 1;
	FOR(i,1,SZ) F0R(j,i+1) 
		scmb[i][j] = scmb[i-1][j]+(j?scmb[i-1][j-1]:0);
}


/**
 * Description: pre-compute factorial mod inverses,
 	* assumes $MOD$ is prime and $SZ < MOD$.
 * Time: O(SZ)
 * Source: KACTL
 * Verification: https://dmoj.ca/problem/tle17c4p5
 */

vi invs, fac, ifac; // make sure to convert to LL before doing any multiplications ...
void genFac(int SZ) {
	invs.rsz(SZ), fac.rsz(SZ), ifac.rsz(SZ); 
	invs[1] = fac[0] = ifac[0] = 1; 
	FOR(i,2,SZ) invs[i] = MOD-(ll)MOD/i*invs[MOD%i]%MOD;
	FOR(i,1,SZ) {
		fac[i] = (ll)fac[i-1]*i%MOD;
		ifac[i] = (ll)ifac[i-1]*invs[i]%MOD;
	}
}

mi comb(int a, int b) {
	if (a < b || b < 0) return 0;
	return mi((ll)fac[a]*ifac[b]%MOD*ifac[a-b]%MOD);
}

int n, m;
int counter;  //even b minus odd b
int evens, odds; //even/odd ?

mi solveNaive(){
	mi ans = 0;

	if((abs(counter) % 2) == 0){
		int hi = (evens+odds)+((evens+odds)&1);
		for(int i = -hi; i <= hi; i+=2){
			ans+=mi(abs(counter+i))*mi(comb(evens+odds, odds+i));
		}
	}
	else{
		int hi = (evens+odds)+((evens+odds)&1);
		for(int i = -hi-2; i <= hi+2; i+=2){
			ans+=mi(abs(counter+1+i))*mi(comb(evens+odds, odds+1+i));
		}
	}
	ans/=2;
	ans/=pow(mi(2), evens+odds-1);
	return ans;
}

vector<pair<pi, mi>> lastpsums;
vector<pair<pi, mi>> curpsums;

mi naivePrefSumComb(int c, int x){
	mi ans = 0;
	for(int i = 0; i <= x; i++){
		ans+=comb(c, i);
	}
	return ans;
}

mi prefSumComb(int c, int x){ //c choose 0 + c choose 1 + ... c choose x
	 assert(c >= 0);
	 if(x < 0) return mi(0);
	 ckmin(x, c);

	 mi ans = 0;

	 pi buildans = mp(x, -1); //operations needed, index
	 for(int i = 0; i < sz(lastpsums); i++){
	 	pi coor = lastpsums[i].f;
	 	ckmin(buildans, mp(abs(coor.f-c)+abs(coor.s-x), i));
	 }

	 if(buildans.s == -1){
	 	ans = naivePrefSumComb(c, x);
	 }
	 else{
	 	ans = lastpsums[buildans.s].s;
	 	pi curcoor = lastpsums[buildans.s].f;
	 	while(curcoor.f < c){
	 		ans = 2*ans-comb(curcoor.f, curcoor.s);
	 		curcoor.f++;
	 	}
	 	while(curcoor.f > c){
	 		ans = invs[2]*ans+invs[2]*comb(curcoor.f-1, curcoor.s);
	 		curcoor.f--;
	 	}
	 	while(curcoor.s < x){
	 		ans+=comb(curcoor.f, curcoor.s+1);
	 		curcoor.s++;
	 	}
	 	while(curcoor.s > x){
	 		ans-=comb(curcoor.f, curcoor.s);
	 		curcoor.s--;
	 	}
	 }

	 // mi naiveans = naivePrefSumComb(c, x);

	 // if(ans != naiveans){
	 // 	dbg(lastpsums[buildans.s]);
	 // }
	 
	 // dbg(c, x, ans, naiveans);

	curpsums.pb(mp(mp(c, x), ans));

	return ans;
}

mi halfComb(int c, int x){ //sum of c choose x+i for even i >= 0
	assert(c >= 0);
	if(c == 0){
		if(x > c) return mi(0);
		if((c-x) % 2 == 0) return mi(1);
		return mi(0);
	}
	mi ans = prefSumComb(c-1, c-x);
	return ans;
}

mi linComb(int c, int x){ //sum of i * (c choose (x+i)) for even i >= 0
	mi ans = 0;
 	ans+=mi(c)*halfComb(c-1, x-1);
 	ans-=mi(x)*halfComb(c, x);
 	return ans;
}

mi solveReg(){
	mi ans = 0;
	ans+=linComb(evens+odds, odds-counter);
	ans+=linComb(evens+odds, evens+counter);
	ans/=mi(2);

	swap(lastpsums, curpsums);
	curpsums.clear();
	ans/=pow(mi(2), evens+odds-1);
	return ans;
}

int main() {
	setIO();
	genFac(600005);
	cin >> n >> m;
	string s;
	cin >> s;

	for(int i = 0; i < n; i++){
		if(s[i] == 'b'){
			if(i % 2 == 0){
				counter++;
			}
			else{
				counter--;
			}
		}
		else if(s[i] == '?'){
			if(i % 2 == 0){
				evens++;
			}
			else{
				odds++;
			}
		}
	}

	mi ans = solveReg();
	ps(ans);

	for(int i = 0; i < m; i++){
		int pos;
		char c;
		cin >> pos >> c;
		pos--;
		if(s[pos] == 'b'){
			if(pos % 2 == 0){
				counter--;
			}
			else{
				counter++;
			}
		}
		else if(s[pos] == '?'){
			if(pos % 2 == 0){
				evens--;
			}
			else{
				odds--;
			}
		}

		s[pos] = c;
		if(s[pos] == 'b'){
			if(pos % 2 == 0){
				counter++;
			}
			else{
				counter--;
			}
		}
		else if(s[pos] == '?'){
			if(pos % 2 == 0){
				evens++;
			}
			else{
				odds++;
			}
		}
		ans = solveReg();
		ps(ans);
	}

	

	// you should actually read the stuff at the bottom
}

/* stuff you should look for
	* int overflow, array bounds
	* special cases (n=1?)
	* do smth instead of nothing and stay organized
	* WRITE STUFF DOWN
*/