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 long long mod = 998244353;
long long n, nr[100002], ans[100002], x, y;
long long logpow(long long a, long long b) {
  if (b < 0) return 0;
  long long p = 1;
  while (b) {
    if (b & 1) p = 1ll * p * a % mod;
    b >>= 1;
    a = 1ll * a * a % mod;
  }
  return p;
}
int main() {
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> x >> y;
    nr[x] += y;
  }
  for (long long i = 100000; i >= 1; i--) {
    long long sq = 0, sum = 0, fr = 0;
    for (long long j = i; j <= 100000; j += i) {
      sq += 1ll * nr[j] % mod * j % mod * j % mod;
      sq %= mod;
      sum += nr[j] * j % mod;
      sum %= mod;
      fr += nr[j];
    }
    sq %= mod;
    sum = 1ll * sum * sum % mod;
    long long x = (sum - sq + mod) % mod;
    ans[i] =
        sq * logpow(2, fr - 2) % mod * ((fr - 1) % mod) +
        x * (logpow(2, fr - 3) * ((fr - 2) % mod) % mod + logpow(2, fr - 2)) %
            mod;
    ans[i] %= mod;
    for (long long j = 2 * i; j <= 100000; j += i)
      ans[i] = (ans[i] - ans[j] + mod) % mod;
  }
  cout << ans[1];
  return 0;
}
