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;
const int N = 1e5 + 10;
int n, x, mu[N];
long long a[N];
long long Mod(long long a, long long b) {
  if (!b) return 1;
  if (b % 2 == 1) return Mod(a, b - 1) * a % 998244353;
  long long z = Mod(a, b >> 1);
  return z * z % 998244353;
}
int main() {
  cin >> n;
  while (n--) cin >> x, cin >> a[x];
  mu[1] = 1;
  for (int i = 1; i < N; i++)
    for (int j = i * 2; j < N; j += i) mu[j] -= mu[i];
  long long ans = 0;
  for (int i = 1; i < N; i++) {
    long long sl = 0, s = 0, sq = 0;
    for (int j = i; j < N; j += i) {
      sl += a[j];
      s = (s + a[j] * j) % 998244353;
      sq = (sq + a[j] * j % 998244353 * j) % 998244353;
    }
    if (sl < 2) continue;
    long long t1 =
        sq * Mod(2, sl - 2) % 998244353 * ((sl - 1) % 998244353) % 998244353;
    long long n2 = (s * s + 998244353 - sq) % 998244353;
    long long t2 = Mod(2, sl - 2);
    if (sl >= 3)
      t2 = (t2 + ((sl - 2) % 998244353) * Mod(2, sl - 3)) % 998244353;
    t2 = t2 * n2 % 998244353;
    ans = (ans + (t1 + t2) % 998244353 * mu[i] + 998244353) % 998244353;
  }
  cout << ans;
  return 0;
}
