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 P = 998244353;
long long power(long long a, long long b) {
  long long res = 1;
  for (; b > 0; b /= 2, a = a * a % P)
    if (b & 1) res = res * a % P;
  return res;
}
int n;
long long cnt[100001], dp[100001];
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  while (n--) {
    int x, f;
    cin >> x >> f;
    cnt[x] = f;
  }
  for (int i = 100000; 1 <= i; i--) {
    auto add = [&](long long &a, long long b) { a = (a + b) % P; };
    long long s = 0;
    long long same = 0;
    long long c = 0;
    for (long long j = i; j <= 100000; j += i) {
      c += cnt[j];
      add(s, j * cnt[j]);
      add(same, j * j % P * cnt[j]);
      if (j > i) add(dp[i], P - dp[j]);
    }
    long long d = c % P;
    add(dp[i], same * (d - 1 + P) % P * power(2, c - 2) % P);
    add(dp[i], (s * s % P - same + P) *
                   ((d - 2 + P) * power(2, c - 3) % P + power(2, c - 2)));
  }
  cout << dp[1] << endl;
  return 0;
}
