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>
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
using namespace std;
typedef long long int ll;
typedef pair<int,int> P;
#define yn {puts("Yes");}else{puts("No");}

const int MAX_N = 1 << 19;

// セグメント木を持つグローバル配列
int nn, dat[2 * MAX_N - 1];

// 初期化
void init(int n_){
    // n_が小さいときの変な挙動を防ぐ
    if(n_ < 10)n_ = 10;
    // 簡単のため要素数を2のべき乗に
    nn = 1;
    while(nn < n_)nn *= 2;

    // 全ての値をINT_MAXに
    for(int i = 0; i < 2 * nn - 1; i++)dat[i] = INT_MAX;
}

// k番目の値(0-indexed)をaに変更
void update(int k, int a){
    // 葉の接点
    k += nn - 1;
    dat[k] = a;
    // 登りながら更新
    while(k > 0){
        k = (k - 1) / 2;
        dat[k] = min(dat[k * 2 + 1], dat[k * 2 + 2]);
    }
}

// [a,b)の最小値を求める
// 後ろのほうの引数は、計算の簡単のための引数。
// kは節点の番号、l, rはその節点が[l, r)に対応づいていることを表す。
// したがって、外からはquery(a, b, 0, 0, nn)として呼ぶ。
int query(int a, int b, int k, int l, int r){
    // [a, b)と[l, r)が交差しなければ、INT_MAX
    if (r <= a || b <= l)return INT_MAX;

    // [a, b)が[l, r)を完全に含んでいれば、この節点の値
    if (a <= l && r <= b)return dat[k];
    else {
        // そうでなければ、2つの子の最小値
        int vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
        int vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
        return min(vl, vr);
    }
}

const ll MOD = 998244353;

int main() {
    int n;
    cin >> n;
    vector<P> v;
    rep(i,n){
        int x, d;
        cin >> x >> d;
        v.push_back(P(x,d));
    }
    v.push_back(P(2000000007, 1));

    sort(v.begin(), v.end());

    init(n+1);

    int x[n+1] = {};
    rep(i,n+1)x[i] = v[i].first;

    update(n, -n);
    drep(i,n){
        int ite = lower_bound(x, x+n+1, v[i].first + v[i].second) - x;
        int tmp = query(i,ite,0,0,nn);
        if(tmp > 1234567)tmp = -i;
        update(i, tmp);
    }

    ll dp[n+1] = {};
    dp[0] = 1;
    rep(i,n){
        dp[i+1] += dp[i];
        dp[i+1] %= MOD;
        int tmp = -query(i,i+1,0,0,nn);
        dp[tmp+1] += dp[i];
        dp[tmp+1] %= MOD;
    }

    cout << dp[n] << endl;
    return 0;
}