#include<bits/stdc++.h>
using namespace std;
#define int long long
const int mod = 998244353;
int a[200005],b[200005];
int posa[400005];
int la[200005],ra[200005],lb[200005],rb[200005];
vector<int> rra[200005],rrb[200005];
int q,op;int n,m;
int s[500005],x[500005];
int nown,nowm;
vector<int>p,pp;int tot;
struct node{
	int l,r;
	int kn,bn;//a
	int km,bm;//b
	int v_mn,v_n,v_m,v_1;			
}tree[2000005];
void push_up(int k){
	int ls = k<<1,rs = k<<1|1;
	tree[k].kn = (tree[ls].kn+tree[rs].kn)%mod;
	tree[k].km = (tree[ls].km+tree[rs].km)%mod;
	tree[k].bn = (tree[ls].bn+tree[rs].bn)%mod;
	tree[k].bm = (tree[ls].bm+tree[rs].bm)%mod;
	tree[k].v_mn = (tree[ls].v_mn+tree[rs].v_mn+tree[ls].km*tree[rs].kn)%mod;
	tree[k].v_n = (tree[ls].v_n+tree[rs].v_n+tree[ls].bm*tree[rs].kn)%mod;
	tree[k].v_m = (tree[ls].v_m+tree[rs].v_m+tree[ls].km*tree[rs].bn)%mod;
	tree[k].v_1 = (tree[ls].v_1+tree[rs].v_1+tree[ls].bm*tree[rs].bn)%mod;
}void build(int l,int r,int k){
	tree[k].l = l,tree[k].r = r;
	if(l+1 == r)return;
	build(l,l+r>>1,k<<1),build(l+r>>1,r,k<<1|1);
}void update(int k,int pos,int op,int k1,int b1){
	int l = tree[k].l,r = tree[k].r;
	if(l+1 == r){
		if(op == 0)tree[k].kn = k1,tree[k].bn = b1;
		else tree[k].km = k1,tree[k].bm = b1;return;
	}int mid = l+r>>1;
	if(pos<mid)update(k<<1,pos,op,k1,b1);
	else update(k<<1|1,pos,op,k1,b1);
	push_up(k);return;
}int query(){
//	cout<<tree[1].kn<<' '<<tree[1].km<<' '<<tree[1].bm<<' '<< tree[1].bn<<endl;
//	cout<<tree[1].v_1<< ' '<<tree[1].v_n <<' '<<tree[1].v_m<<' '<<tree[1].v_mn<<endl;
	int res = (nown*nowm%mod*tree[1].v_mn+nown*tree[1].v_n+nowm*tree[1].v_m+tree[1].v_1)%mod;return (res+mod)%mod;}
signed main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin >>q >>op;
	cin >>a[++n]>>b[++m];b[1] = -b[1];
	p.push_back(b[1]),p.push_back(a[1]),pp.push_back(b[1]),pp.push_back(a[1]);
	for(int i = 1;i<=q;i++){
		cin>>s[i]>>x[i];
		if(s[i] == 1)x[i] = -x[i];
		p.push_back(x[i]),pp.push_back(x[i]);
	}//cout<<111<<endl;
	sort(p.begin(),p.end());sort(pp.begin(),pp.end());p.erase(unique(p.begin(),p.end()),p.end());
	for(int i= 0;i<p.size();i++){
	//	cout<<i<<endl;
		int pos = upper_bound(pp.begin(),pp.end(),p[i])-pp.begin();posa[i] = pos;
	}for(int i = 1;i<=q;i++){
		if(s[i] == 0)a[++n] = x[i];
		else b[++m] = x[i];
	}//cout<<111<<endl;
	for(int i = 1;i<=n;i++){
		int ppp = lower_bound(p.begin(),p.end(),a[i])-p.begin();
	//	cout<<ppp<<endl;
		a[i] = posa[ppp]--;
	}for(int i = 1;i<=m;i++){
		int ppp = lower_bound(p.begin(),p.end(),b[i])-p.begin();
	//	cout<<ppp<< endl;
		b[i] = posa[ppp]--;
	}vector<int>p;
	for(int i = 1;i<=n;i++)ra[i] = n+1;
	for(int i = 1;i<=m;i++)rb[i] = m+1;
	for(int i = 1;i<=n;i++){
		while(!p.empty() and a[i]<a[p.back()])p.pop_back();
		if(!p.empty())la[i] = p.back();p.push_back(i);
	}p.clear();
	for(int i = n;i>=1;i--){
		while(!p.empty() and a[i]<a[p.back()])p.pop_back();
		if(!p.empty())ra[i] = p.back(),rra[p.back()].push_back(i);p.push_back(i);
	}p.clear();
	for(int i = 1;i<=m;i++){
		while(!p.empty() and b[i]>b[p.back()])p.pop_back();
		if(!p.empty())lb[i] = p.back();p.push_back(i);
	}p.clear();
	for(int i = m;i>=1;i--){
		while(!p.empty() and b[i]>b[p.back()])p.pop_back();
		if(!p.empty())rb[i] = p.back(),rrb[p.back()].push_back(i);;p.push_back(i);
	}p.clear();
//	for(int i = 1;i<=n;i++)cout<<a[i]<<' ';cout<<endl;
//	for(int i = 1;i<=m;i++)cout<<b[i]<<' ';cout<<endl;
//	for(int i = 1;i<=m;i++)cout<<lb[i]<<' ';cout<<endl;
//	for(int i = 1;i<=m;i++)cout<<rb[i]<< ' ';cout<<endl;
	tot = n+m;build(1,m+n+1,1);
	update(1,a[1],0,1,0);update(1,b[1],1,1,0);nown = 1,nowm = 1;
	//cout<<query()<<endl;
	for(int i = 1;i<=q;i++){
		if(s[i] == 0){
			nown++;update(1,a[nown],0,nown-la[nown],-(nown-1)*(nown-la[nown]));
			for(int x:rra[nown]){
				update(1,a[x],0,0,(x-la[x])*(ra[x]-x));
			}
		}else{
			nowm++;update(1,b[nowm],1,nowm-lb[nowm],-(nowm-1)*(nowm-lb[nowm]));
			for(int x:rrb[nowm]){
		//		cout << ' '<<x <<' '<<(x-lb[x])*(rb[x]-x)<<endl;
				update(1,b[x],1,0,(x-lb[x])*(rb[x]-x));
			}
		}if(op == 1)cout<<query()<< endl;
	}if(op == 0)cout<<query()<<endl;
	return 0;
}
/*
我们考虑把全黑矩阵转化为 min_{l1,r1}(ai)>=max_{l2,r2}(-bi) 的区间个数
考虑单调栈，维护 ai 在哪里作为最小值，模拟单调栈的过程
发现对于一个序列的答案容易表示成 kn+b
上线段树维护一下
2 1  
3 3
1 1
1 2

3 1 2
-3
*/