Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
template <typename Elem>
int quick_power(int a, Elem b, int Mod) {
  int ans = 1;
  while (b) {
    if (b & 1) {
      ans = 1ll * ans * a % Mod;
    }
    b >>= 1;
    a = 1ll * a * a % Mod;
  }
  return ans;
}
const int Maxn = 100000;
const int Mod = 998244353;
int n;
long long sum_0[Maxn + 5];
int sum_1[Maxn + 5], sum_2[Maxn + 5], sum_3[Maxn + 5];
int f[Maxn + 5];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x, a;
    scanf("%d%d", &x, &a);
    int s_1 = 1ll * x * a % Mod, s_2 = 1ll * x * x % Mod * a % Mod,
        s_3 = 1ll * x * x % Mod * a % Mod * (a - 1) % Mod;
    for (int j = 1; j * j <= x; j++) {
      if (x % j == 0) {
        sum_3[j] = (sum_3[j] + s_3 + 2ll * s_1 * sum_1[j]) % Mod;
        sum_2[j] = (sum_2[j] + s_2) % Mod;
        sum_1[j] = (sum_1[j] + s_1) % Mod;
        sum_0[j] = (sum_0[j] + a);
        if (j * j != x) {
          sum_3[x / j] = (sum_3[x / j] + s_3 + 2ll * s_1 * sum_1[x / j]) % Mod;
          sum_2[x / j] = (sum_2[x / j] + s_2) % Mod;
          sum_1[x / j] = (sum_1[x / j] + s_1) % Mod;
          sum_0[x / j] = (sum_0[x / j] + a);
        }
      }
    }
  }
  for (int i = Maxn; i > 0; i--) {
    if (sum_0[i] == 0) {
      continue;
    }
    int p_2 = 0, p_3 = 0;
    if (sum_0[i] > 2) {
      p_3 = quick_power(2, sum_0[i] - 3, Mod);
      p_2 = (p_3 << 1) % Mod;
    } else if (sum_0[i] > 1) {
      p_2 = quick_power(2, sum_0[i] - 2, Mod);
    }
    if (sum_0[i] > 1) {
      f[i] = (f[i] + 1ll * p_2 * ((sum_0[i] - 1) % Mod) % Mod * sum_2[i]) % Mod;
      f[i] = (f[i] + 1ll * p_2 * sum_3[i]) % Mod;
    }
    if (sum_0[i] > 2) {
      f[i] = (f[i] + 1ll * p_3 * ((sum_0[i] - 2) % Mod) % Mod * sum_3[i]) % Mod;
    }
    for (int j = i + i; j <= Maxn; j += i) {
      f[i] = (f[i] - f[j] + Mod) % Mod;
    }
  }
  printf("%d\n", f[1]);
  return 0;
}