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)

/*
　　　∫ ∫ ∫
　　　ノヽ
　　（＿　 ）
　（＿　　　 ）
（＿＿＿＿＿＿ ）
　ヽ(´･ω･)ﾉ　
　　 |　 /
　　 UU
*/
#pragma region macro
#include <bits/stdc++.h>
typedef long long int64;
using namespace std;
using P = pair<int64, int64>;
typedef vector<int> vi;
const int MOD = (int)1e9 + 7;
const int64 INF = 1LL << 62;
const int inf = 1<<30;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
#define REP(i, n) for (int i = 0; i < (n); i++)
#define FOR(i,s,n) for (int i = s; i < (n); i++)
#define ALL(obj) (obj).begin(), (obj).end() //コンテナじゃないと使えない!!
#define debug(x) cerr << #x << ": " << x << "\n";
#define mp make_pair
#define bn '\n'
template <typename T>
ostream& operator<<(ostream& os, const vector<T> &V){
    int N = V.size();
    REP(i,N){
        os << V[i];
        if (i!=N-1) os << " ";
    }
    os << "\n";
    return os;
}
template <typename T,typename S>
ostream& operator<<(ostream& os, pair<T,S> const&P){
    os << "(";
    os << P.first;
    os << " , ";
    os << P.second;
    os << ")";
    return os;
}
template <typename T>
ostream& operator<<(ostream& os, set<T> &S){
    auto it=S.begin();
    while(it!=S.end()){
        os << *it;
        os << " ";
        it++;
    }
    os << "\n";
    return os;
}
template <typename T>
ostream& operator<<(ostream& os, deque<T> &q){
    for(auto it=q.begin();it<q.end();it++){
        os<<*it;
        os<<" ";
    }
     os<<endl;
    return os;
}
vector<pair<int,int>> dxdy = {mp(0,1),mp(1,0),mp(-1,0),mp(0,-1)};
#pragma endregion
//fixed<<setprecision(10)<<ans<<endl;

template <typename T>
struct SegmentTree{
    /*
    ~~~~1-indexで実装~~~~
            1
        2       3
      4   5   6    7

    ~~~~親・兄弟・子へのアクセスの仕方~~~~
                i<<1
            i           i^1
    i<<1|0   i<<1|1
    */


    using F = function<T(T,T)>;
    int N;
    F func; //関数(minとか)
    T identity; //単位元
    vector<T> data; //上から添字 2*Nくらいのノード
    SegmentTree(){}
    SegmentTree(F f,T identity):func(f),identity(identity){}

    void init(int n_){
        N=1;
        while(N<n_) N<<=1; //完全二分木がいいので2^k個にする
        data.assign(N<<1, identity); //2N個のノード
    }

    //木を構成(vectorを元に各ノードの値を計算)
    void build(const vector<T>& v){
        int n_ = v.size();
        init(n_);
        for(int i=0; i<n_; i++){
            data[N+i] = v[i]; //葉
        }
        for(int i=N-1; i; i--){
            data[i] = func(data[i<<1], data[(i<<1)+1] ); //子をみて親を更新
        }
    }

    //k番目の値をxに変える
    void set_val(int k, T x){
        k+=N; //indexをセグ木でのindexに変換
        data[k] = x; //値の書き換え(葉)
        while(k>>=1){ //右シフトして0にならない間
            data[k] = func(data[k<<1], data[(k<<1)+1] );
        }
    }

    //https://hcpc-hokudai.github.io/archive/structure_segtree_001.pdf
    //閉開区間[left,right)で
    T query(int left,int right){
        T left_val=identity, right_val=identity; 
        for(int l=left+N,r=right+N; l<r; l>>=1,r>>=1){
            if(l&1) left_val = func(left_val, data[l++] );
            if(r&1) right_val = func(data[--r], right_val);
        }
        return func(left_val, right_val);
    }

    //葉の値を取得する
    T get_val(int idx){
        return data[idx+N];
    }
};
template <typename T>
ostream& operator<<(ostream& os, SegmentTree<T> &S){
    int N = S.data.size();
    int cnt = 0;
    int up = 1;
    for(int i=1;i<N;i++){
        if(S.data[i]!=S.identity){
            os << S.data[i];
        }else os << "単";
        os << " ";
        cnt++;
        if (cnt == up){
            os << "\n\n";
            up<<=1;
            cnt=0;
        }else if (cnt%2==0) os<<",";
    }
    os << "\n";
    return os;
};

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N;
    cin >> N;
    vector<pair<int64,int64>> XD;
    int64 x,d;
    REP(i,N){
        cin >> x >> d;
        XD.emplace_back(x,d);
    }

    sort(ALL(XD));  
    XD.emplace_back(1E11, INF); //番兵

    SegmentTree<int64> Seg(
        [](int64 a,int64 b)->int64{
            return max(a,b);
        }, -INF
    );

    vector<int64> XD_sum;
    REP(i,N){
        XD_sum.emplace_back(XD[i].first+XD[i].second);
    }

    Seg.init(N);
    Seg.build(XD_sum);

    vector<int64> DP(N+1,0);
    DP[N]=1;
    REP(i,N){
        int idx = N-i-1;
        DP[idx] += DP[idx+1]; //idx番目を起動しない
        int64 x = Seg.query(idx, lower_bound(ALL(XD), mp(XD[idx].second + XD[idx].first , 0LL)) - XD.begin()); //idx番目が動いたときに起動するやつの中で一番遠くまで起動させる奴
        Seg.set_val(idx, x);

        int right_ = lower_bound(ALL(XD), mp(x,0LL)) - XD.begin();
        DP[idx] += DP[right_]; //idx番目を起動したときに勝手に起動されないロビットの一番左から右端の場合の数
        DP[idx]%=998244353;
    }

    auto ans=DP[0];



    cout << ans << endl;
}