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 MOD = 119 << 23 | 1;
const int FFTMOD = 119 << 23 | 1;
const int INF = (int)1e9 + 23111992;
const long long LINF = (long long)1e18 + 23111992;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T>
inline int chkmin(T& a, const T& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T>
inline int chkmax(T& a, const T& val) {
  return a < val ? a = val, 1 : 0;
}
inline unsigned long long isqrt(unsigned long long k) {
  unsigned long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
inline int sign(long double x, long double y) { return sign(x - y); }
mt19937 mt(chrono::high_resolution_clock::now().time_since_epoch().count());
inline int mrand() { return abs((int)mt()); }
inline int mrand(int k) { return abs((int)mt()) % k; }
void chemthan() {
  int m;
  cin >> m;
  vector<int> f(1e5 + 5);
  for (int i = (0); i < (m); ++i) {
    int a;
    cin >> a;
    cin >> f[a];
  }
  auto calc = [&](int d) {
    int sum = 0, sq_sum = 0;
    long long k = 0;
    for (int i = d; i < int((f).size()); i += d) {
      addmod(sum, mult(f[i], i));
      addmod(sq_sum, mult(f[i], mult(i, i)));
      k += f[i];
    }
    if (k < 2) return 0;
    int cross_sum = mult(sum, sum);
    submod(cross_sum, sq_sum);
    cross_sum = mult(cross_sum, inv(2));
    int res = mult(sq_sum, mult((k - 1) % MOD, fpow(2, k - 2)));
    addmod(res, mult(cross_sum, mult(k % MOD, fpow(2, k - 2))));
    return res;
  };
  auto mob = [&](int d) {
    map<int, int> hs;
    for (int i = 2; i * i <= d; i++) {
      while (d % i == 0) hs[i]++, d /= i;
    }
    if (1 < d) hs[d]++;
    int res = 1;
    for (auto [p, c] : hs) {
      if (1 < c) return 0;
      res *= -1;
    }
    return res;
  };
  int res = 0;
  for (int d = (1); d < (int((f).size())); ++d) {
    addmod(res, mult(mob(d) + MOD, calc(d)));
  }
  cout << res << "\n";
}
int32_t main(int32_t argc, char* argv[]) {
  ios_base::sync_with_stdio(0), cin.tie(0);
  if (argc > 1) {
    assert(freopen(argv[1], "r", stdin));
  }
  if (argc > 2) {
    assert(freopen(argv[2], "wb", stdout));
  }
  chemthan();
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}
