Problem:
Score : 600 points

Problem Statement

There are N robots numbered 1 to N placed on a number line. Robot i is placed at coordinate X_i. When activated, it will travel the distance of D_i in the positive direction, and then it will be removed from the number line. All the robots move at the same speed, and their sizes are ignorable.

Takahashi, who is a mischievous boy, can do the following operation any number of times (possibly zero) as long as there is a robot remaining on the number line.

Choose a robot and activate it. This operation cannot be done when there is a robot moving.

While Robot i is moving, if it touches another robot j that is remaining in the range [X_i, X_i + D_i) on the number line, Robot j also gets activated and starts moving. This process is repeated recursively.

How many possible sets of robots remaining on the number line are there after Takahashi does the operation some number of times? Compute this count modulo 998244353, since it can be enormous.

Constraints

1 \leq N \leq 2 \times 10^5

-10^9 \leq X_i \leq 10^9

1 \leq D_i \leq 10^9

X_i \neq X_j (i \neq j)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
X_1 D_1
:
X_N D_N

Output

Print the number of possible sets of robots remaining on the number line, modulo 998244353.

Sample Input 1

2
1 5
3 3

Sample Output 1

3

There are three possible sets of robots remaining on the number line: \{1, 2\}, \{1\}, and \{\}.

These can be achieved as follows:

If Takahashi activates nothing, the robots \{1, 2\} will remain.

If Takahashi activates Robot 1, it will activate Robot 2 while moving, after which there will be no robots on the number line. This state can also be reached by activating Robot 2 and then Robot 1.

If Takahashi activates Robot 2 and finishes doing the operation, the robot \{1\} will remain.

Sample Input 2

3
6 5
-1 10
3 3

Sample Output 2

5

There are five possible sets of robots remaining on the number line: \{1, 2, 3\}, \{1, 2\}, \{2\}, \{2, 3\}, and \{\}.

Sample Input 3

4
7 10
-10 3
4 3
-4 3

Sample Output 3

16

None of the robots influences others.

Sample Input 4

20
-8 1
26 4
0 5
9 1
19 4
22 20
28 27
11 8
-3 20
-25 17
10 4
-18 27
24 28
-11 19
2 27
-2 18
-1 12
-24 29
31 29
29 7

Sample Output 4

110

Remember to print the count modulo 998244353.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;

#define endl '\n'
#define lfs cout<<fixed<<setprecision(10)
#define ALL(a)  (a).begin(),(a).end()
#define ALLR(a)  (a).rbegin(),(a).rend()
#define spa << " " <<
#define fi first
#define se second
#define MP make_pair
#define MT make_tuple
#define PB push_back
#define EB emplace_back
#define rep(i,n,m) for(ll i = (n); i < (ll)(m); i++)
#define rrep(i,n,m) for(ll i = (m) - 1; i >= (ll)(n); i--)
using ll = long long;
using ld = long double;
const ll MOD1 = 1e9+7;
const ll MOD9 = 998244353;
const ll INF = 1e18;
using P = pair<ll, ll>;
template<typename T>
bool chmin(T &a,T b){if(a>b){a=b;return true;}else return false;}
template<typename T>
bool chmax(T &a,T b){if(a<b){a=b;return true;}else return false;}
ll median(ll a,ll b, ll c){return a+b+c-max({a,b,c})-min({a,b,c});}
void ans1(bool x){if(x) cout<<"Yes"<<endl;else cout<<"No"<<endl;}
void ans2(bool x){if(x) cout<<"YES"<<endl;else cout<<"NO"<<endl;}
void ans3(bool x){if(x) cout<<"Yay!"<<endl;else cout<<":("<<endl;}
template<typename T1,typename T2>
void ans(bool x,T1 y,T2 z){if(x)cout<<y<<endl;else cout<<z<<endl;}  
template<typename T>
void debug(vector<vector<T>>&v,ll h,ll w){for(ll i=0;i<h;i++)
{cout<<v[i][0];for(ll j=1;j<w;j++)cout spa v[i][j];cout<<endl;}};
void debug(vector<string>&v,ll h,ll w){for(ll i=0;i<h;i++)
{for(ll j=0;j<w;j++)cout<<v[i][j];cout<<endl;}};
template<typename T>
void debug(vector<T>&v,ll n){if(n!=0)cout<<v[0];
for(ll i=1;i<n;i++)cout spa v[i];cout<<endl;};
template<typename T>
vector<vector<T>>vec(ll x, ll y, T w){
  vector<vector<T>>v(x,vector<T>(y,w));return v;}
ll gcd(ll x,ll y){ll r;while(y!=0&&(r=x%y)!=0){x=y;y=r;}return y==0?x:y;}
vector<ll>dx={1,0,-1,0,1,1,-1,-1};
vector<ll>dy={0,1,0,-1,1,-1,1,-1};
template<typename T>
vector<T> make_v(size_t a,T b){return vector<T>(a,b);}
template<typename... Ts>
auto make_v(size_t a,Ts... ts){
  return vector<decltype(make_v(ts...))>(a,make_v(ts...));
}
ostream &operator<<(ostream &os, pair<ll, ll>&p){
  return os << p.first << " " << p.second;
}  
template< int mod >
struct ModInt {
  ll x;

  ModInt() : x(0) {}

  ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}

  ModInt &operator+=(const ModInt &p) {
    if((x += p.x) >= mod) x -= mod;
    return *this;
  }

  ModInt &operator-=(const ModInt &p) {
    if((x += mod - p.x) >= mod) x -= mod;
    return *this;
  }

  ModInt &operator*=(const ModInt &p) {
    x = (int) (1LL * x * p.x % mod);
    return *this;
  }

  ModInt &operator/=(const ModInt &p) {
    *this *= p.inverse();
    return *this;
  }

  ModInt operator-() const { return ModInt(-x); }

  ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }

  ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }

  ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }

  ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }

  bool operator==(const ModInt &p) const { return x == p.x; }

  bool operator!=(const ModInt &p) const { return x != p.x; }

  ModInt inverse() const {
    int a = x, b = mod, u = 1, v = 0, t;
    while(b > 0) {
      t = a / b;
      swap(a -= t * b, b);
      swap(u -= t * v, v);
    }
    return ModInt(u);
  }

  ModInt pow(int64_t n) const {
    ModInt ret(1), mul(x);
    while(n > 0) {
      if(n & 1) ret *= mul;
      mul *= mul;
      n >>= 1;
    }
    return ret;
  }

  friend ostream &operator<<(ostream &os, const ModInt &p) {
    return os << p.x;
  }

  friend istream &operator>>(istream &is, ModInt &a) {
    int64_t t;
    is >> t;
    a = ModInt< mod >(t);
    return (is);
  }

  static int get_mod() { return mod; }
};
using modint = ModInt< MOD9 >;
modint pow2(int64_t n, int64_t x){
    return modint(n).pow(x);
}
template<typename T>
struct SegmentTree{
  using F = function<T(T,T)>;
  vector<T> data;
  ll n,lastlen = 1;
  T func(T a, T b){return a > b ? a : b;};
  bool check (T a, T b){return a <= b;};
  T iden = -INT_MAX; //単位元を変えるのを忘れない！！！！！！！！！！
  SegmentTree(vector<T> &v){
    build(v);
  }
  SegmentTree(ll size){
    n = size;
    while(lastlen < n)lastlen *= 2;
    data.assign(lastlen*2-1,iden);
  }
  void build(vector<T> &v){
    n = (ll)v.size();
    while(lastlen < n)lastlen *= 2;
    data.assign(lastlen*2-1,iden);
    for(ll i=0;i<n;i++)data[i+lastlen-1] = v[i];
    for(ll i=lastlen-2;i>=0;i--){
      data[i] = func(data[2*i+1], data[2*i+2]);
    }
  }
  void build(){
    for(ll i=lastlen-2;i>=0;i--){
      data[i] = func(data[2*i+1], data[2*i+2]);
    }
  }
  void update(ll point, T x){
    point+=lastlen-1;
    data[point] = x;
    while(point!=0){
      point=(point-1)/2;
      data[point]=func(data[2*point+1],data[2*point+2]);
    }
  }
  T query(ll l,ll r){
    T left = iden, right = iden;
    l+=lastlen-1, r+=lastlen-1;
    while(l != r){
      if(~l&1)left = func(left, data[l++]);
      if(~r&1)right = func(data[--r], right);
      l >>= 1,r >>= 1;
    }
    return func(left, right);
  }
  T &operator[](ll x){
    return data[lastlen-1+x];
  }
  void print(){
    for(ll i=0;i<n;i++){
      if(i!=0)cout<<" ";
      cout<<data[lastlen-1+i];
    }
    cout<<endl;
  }
  T lower_bound(ll p,T x){
    ll sumbuf=iden, l=p,r=p+1;
    p = p + lastlen - 1;
    while(1){
      if(!check(func(sumbuf,data[p]), x)){
        if(p%2==1){
          p = (p-1)/2;
          r = l + (r-l)*2;
        }
        else if(r>=n){
          return n;
        }
        else{
          func(sumbuf,data[p]);
          p >>= 1;
          ll x = r + (r-l)*2; 
          l = r; r = x;
        }
      }
      else if(p >= lastlen - 1)return r - 1;
      else if(!check(func(sumbuf,data[2*p+1]),x)){
        sumbuf = func(sumbuf, data[2*p+1]);
        p = 2*p+2; l = (r+l)/2;
      }
      else{
        p = 2*p+1; r = (r+l)/2;
      }
    }  
  }
};
//0-indexedで各値が何番目に小さいかを返す

template<typename T>
map<T,ll> compress(vector<T> &v){
  auto t=v;
  sort(ALL(t));
  t.erase(unique(ALL(t)),t.end());
  map<T, ll> ret;
  for(ll i=0;i<(ll)t.size();i++)
    ret[t[i]]=i;
  return ret;
}
int main(){
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  ll res=0,buf=0;
  bool judge = true;
  ll n;cin>>n;
  vector<P>p(n);
  rep(i,0,n)cin>>p[i].fi>>p[i].se;
  vector<ll>val(n);
  rep(i,0,n){
    val[i]=p[i].fi;
  }
  val.PB(INF);
  sort(ALL(p));
  auto mp=compress(val);
  ll sz=mp.size();
  SegmentTree<ll> seg(sz);
  seg[sz-1]=n;
  seg.build();
  rrep(i,0,n){
    auto idx=mp.lower_bound(p[i].fi+p[i].se);
    //cout<<idx->fi spa idx->se<<endl;
    auto q=seg.query(i,idx->se);
    seg.update(i,max(i+1,q));
    //cout<<i spa i+1 spa q<<endl;
  }
  vector<modint>dp(n+1);
  dp[0]=1;
  //seg.print();
  rep(i,0,n){
    dp[seg[i]]+=dp[i];
    dp[i+1]+=dp[i];
  }
  cout<<dp[n]<<endl;
  return 0;
}