You are given a multiset S. Over all pairs of subsets A and B, such that:

  * B ⊂ A; 
  * |B| = |A| - 1; 
  * greatest common divisor of all elements in A is equal to one; 



find the sum of ∑_{x ∈ A}{x} ⋅ ∑_{x ∈ B}{x}, modulo 998 244 353.

Input

The first line contains one integer m (1 ≤ m ≤ 10^5): the number of different values in the multiset S.

Each of the next m lines contains two integers a_i, freq_i (1 ≤ a_i ≤ 10^5, 1 ≤ freq_i ≤ 10^9). Element a_i appears in the multiset S freq_i times. All a_i are different.

Output

Print the required sum, modulo 998 244 353.

Examples

Input


2
1 1
2 1


Output


9


Input


4
1 1
2 1
3 1
6 1


Output


1207


Input


1
1 5


Output


560

Note

A multiset is a set where elements are allowed to coincide. |X| is the cardinality of a set X, the number of elements in it.

A ⊂ B: Set A is a subset of a set B.

In the first example B=\{1\}, A=\{1,2\} and B=\{2\}, A=\{1, 2\} have a product equal to 1⋅3 + 2⋅3=9. Other pairs of A and B don't satisfy the given constraints.
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
const int MOD = 998244353;
const int inf = (1 << 30) - 1;
const ll INF = (1LL << 60) - 1;
template <typename T>
bool chmax(T &x, const T &y) {
  return (x < y) ? (x = y, true) : false;
};
template <typename T>
bool chmin(T &x, const T &y) {
  return (x > y) ? (x = y, true) : false;
};
struct io_setup {
  io_setup() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout << fixed << setprecision(15);
  }
} io_setup;
template <int mod>
struct Mod_Int {
  int x;
  Mod_Int() : x(0) {}
  Mod_Int(ll y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}
  Mod_Int &operator+=(const Mod_Int &p) {
    if ((x += p.x) >= mod) x -= mod;
    return *this;
  }
  Mod_Int &operator-=(const Mod_Int &p) {
    if ((x += mod - p.x) >= mod) x -= mod;
    return *this;
  }
  Mod_Int &operator*=(const Mod_Int &p) {
    x = (int)(1LL * x * p.x % mod);
    return *this;
  }
  Mod_Int &operator/=(const Mod_Int &p) {
    *this *= p.inverse();
    return *this;
  }
  Mod_Int &operator++() { return *this += Mod_Int(1); }
  Mod_Int operator++(int) {
    Mod_Int tmp = *this;
    ++*this;
    return tmp;
  }
  Mod_Int &operator--() { return *this -= Mod_Int(1); }
  Mod_Int operator--(int) {
    Mod_Int tmp = *this;
    --*this;
    return tmp;
  }
  Mod_Int operator-() const { return Mod_Int(-x); }
  Mod_Int operator+(const Mod_Int &p) const { return Mod_Int(*this) += p; }
  Mod_Int operator-(const Mod_Int &p) const { return Mod_Int(*this) -= p; }
  Mod_Int operator*(const Mod_Int &p) const { return Mod_Int(*this) *= p; }
  Mod_Int operator/(const Mod_Int &p) const { return Mod_Int(*this) /= p; }
  bool operator==(const Mod_Int &p) const { return x == p.x; }
  bool operator!=(const Mod_Int &p) const { return x != p.x; }
  Mod_Int inverse() const {
    assert(*this != Mod_Int(0));
    return pow(mod - 2);
  }
  Mod_Int pow(ll k) const {
    Mod_Int now = *this, ret = 1;
    for (; k; k >>= 1, now *= now) {
      if (k & 1) ret *= now;
    }
    return ret;
  }
  friend ostream &operator<<(ostream &os, const Mod_Int &p) {
    return os << p.x;
  }
  friend istream &operator>>(istream &is, Mod_Int &p) {
    ll a;
    is >> a;
    p = Mod_Int<mod>(a);
    return is;
  }
};
using mint = Mod_Int<MOD>;
int main() {
  int N;
  cin >> N;
  int MAX = 100000;
  vector<int> cnt(MAX + 1, 0);
  for (int i = 0; i < N; i++) {
    int a;
    cin >> a;
    cin >> cnt[a];
  }
  mint ans = 0;
  vector<mint> K(MAX + 1, 0);
  for (int i = 1; i <= MAX; i++) {
    mint s1 = 0, s2 = 0;
    ll n = 0;
    mint tmp = -K[i];
    if (i == 1) tmp++;
    for (int j = i; j <= MAX; j += i) {
      K[j] += tmp;
      s1 += mint(j) * cnt[j], s2 += mint(j) * mint(j) * cnt[j], n += cnt[j];
    }
    s1 *= s1, s1 -= s2;
    if (n >= 2) {
      ans += tmp * s2 * mint(n - 1) * mint(2).pow(n - 2);
      ans += tmp * s1 * mint(2).pow(n - 2);
    }
    if (n >= 3) {
      ans += tmp * s1 * mint(n - 2) * mint(2).pow(n - 3);
    }
  }
  cout << ans << '\n';
}
