#include <bits/stdc++.h>
using namespace std;
#define int long long
typedef int ll;
typedef long double ld;
const ll N = 200005;
char en = '\n';
ll inf = 1e16;
ll mod = 998244353;
ll power(ll x, ll n, ll mod) {
  ll res = 1;
  x %= mod;
  while (n) {
    if (n & 1)
      res = (res * x) % mod;
    x = (x * x) % mod;
    n >>= 1;
  }
  return res;
}
ll n;
pair<ll, ll> robots[N];
ll nxt[N];
ll onlyX[N];

//
// Created by cs160 on 09-04-2019.
//
// ashishgup git

struct data {
  // Use required attributes
  int mn;

  // Default Values
  data() : mn(-1e9){};
};

struct SegTree {
  int N;
  vector<data> st;
  vector<bool> cLazy;
  vector<int> lazy;

  void init(int n) {
    N = n;
    st.resize(4 * N + 5);
    cLazy.assign(4 * N + 5, false);
    lazy.assign(4 * N + 5, 0);
  }

  // Write reqd merge functions
  void merge(data &cur, data &l, data &r) { cur.mn = max(l.mn, r.mn); }

  // Handle lazy propagation appriopriately
  void propagate(int node, int L, int R) {
    if (L != R) {
      cLazy[node * 2] = 1;
      cLazy[node * 2 + 1] = 1;
      lazy[node * 2] = lazy[node];
      lazy[node * 2 + 1] = lazy[node];
    }
    st[node].mn = lazy[node];
    cLazy[node] = 0;
  }

  void build(int node, int L, int R) {
    if (L == R) {
      st[node].mn = 1e9;
      return;
    }
    int M = (L + R) / 2;
    build(node * 2, L, M);
    build(node * 2 + 1, M + 1, R);
    merge(st[node], st[node * 2], st[node * 2 + 1]);
  }

  data Query(int node, int L, int R, int i, int j) {
    if (cLazy[node])
      propagate(node, L, R);
    if (j < L || i > R)
      return data();
    if (i <= L && R <= j)
      return st[node];
    int M = (L + R) / 2;
    data left = Query(node * 2, L, M, i, j);
    data right = Query(node * 2 + 1, M + 1, R, i, j);
    data cur;
    merge(cur, left, right);
    return cur;
  }

  data pQuery(int node, int L, int R, int pos) {
    if (cLazy[node])
      propagate(node, L, R);
    if (L == R)
      return st[node];
    int M = (L + R) / 2;
    if (pos <= M)
      return pQuery(node * 2, L, M, pos);
    else
      return pQuery(node * 2 + 1, M + 1, R, pos);
  }

  void Update(int node, int L, int R, int i, int j, int val) {
    if (cLazy[node])
      propagate(node, L, R);
    if (j < L || i > R)
      return;
    if (i <= L && R <= j) {
      cLazy[node] = 1;
      lazy[node] = val;
      propagate(node, L, R);
      return;
    }
    int M = (L + R) / 2;
    Update(node * 2, L, M, i, j, val);
    Update(node * 2 + 1, M + 1, R, i, j, val);
    merge(st[node], st[node * 2], st[node * 2 + 1]);
  }

  void pUpdate(int node, int L, int R, int pos, int val) {
    if (cLazy[node])
      propagate(node, L, R);
    if (L == R) {
      cLazy[node] = 1;
      lazy[node] = val;
      propagate(node, L, R);
      return;
    }
    int M = (L + R) / 2;
    if (pos <= M)
      pUpdate(node * 2, L, M, pos, val);
    else
      pUpdate(node * 2 + 1, M + 1, R, pos, val);
    merge(st[node], st[node * 2], st[node * 2 + 1]);
  }

  data query(int pos) { return pQuery(1, 1, N, pos); }

  data query(int l, int r) {
    if (l > r)
      return data();
    return Query(1, 1, N, l, r);
  }

  void update(int pos, int val) { pUpdate(1, 1, N, pos, val); }

  void update(int l, int r, int val) { Update(1, 1, N, l, r, val); }
};

// Problem 1 (Max Query - Point Update with Coordinate Compression):
// http://codeforces.com/gym/100733/problem/F Solution 1:
// http://codeforces.com/gym/100733/submission/41643795

// Problem 2 (Min Query - Offline processing):
// https://codeforces.com/problemset/problem/522/D Solution 2:
// https://codeforces.com/contest/522/submission/45493164

ll dp[N];

ll recur(ll pos) {
  if (pos == n + 1)
    return 1;
  ll &ans = dp[pos];
  if (ans != -1)
    return ans;

  ans = recur(pos + 1);
  ans += recur(nxt[pos] + 1);

  if (ans >= mod)
    ans -= mod;
  return ans;
}

int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);

  cin >> n;

  for (ll i = 1; i <= n; i++) {
    cin >> robots[i].first >> robots[i].second;
  }

  sort(robots + 1, robots + n + 1);

  SegTree obj;
  obj.init(n);
  for (ll i = 1; i <= n; i++) {
    onlyX[i] = robots[i].first;
    obj.update(i, i);
  }

  for (ll i = n; i >= 1; i--) {
    nxt[i] = i;

    ll l = i;
    ll r = lower_bound(onlyX + i, onlyX + n + 1,
                       robots[i].first + robots[i].second) -
           onlyX;
    r--;
    nxt[i] = max(nxt[i], obj.query(l, r).mn);
    obj.update(i, nxt[i]);
  }

  memset(dp, -1, sizeof(dp));
  cout << recur(1) << en;

  return 0;
}


Measured memory_bytes: 23808