Problem: Two soldiers are playing a game. At the beginning first of them chooses a positive integer n and gives it to the second soldier. Then the second one tries to make maximum possible number of rounds. Each round consists of choosing a positive integer x > 1, such that n is divisible by x and replacing n with n / x. When n becomes equal to 1 and there is no more possible valid moves the game is over and the score of the second soldier is equal to the number of rounds he performed.

To make the game more interesting, first soldier chooses n of form a! / b! for some positive integer a and b (a ≥ b). Here by k! we denote the factorial of k that is defined as a product of all positive integers not large than k.

What is the maximum possible score of the second soldier?

Input

First line of input consists of single integer t (1 ≤ t ≤ 1 000 000) denoting number of games soldiers play.

Then follow t lines, each contains pair of integers a and b (1 ≤ b ≤ a ≤ 5 000 000) defining the value of n for a game.

Output

For each game output a maximum score that the second soldier can get.

Examples

Input

2
3 1
6 3


Output

2
5

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:2000000")
#pragma comment(linker, "/HEAP:2000000")
using namespace std;
void print_width(long long x) {
  std::cout << std::fixed;
  std::cout << std::setprecision(x);
}
long long power(long long x, long long y, long long p = 1000000007) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void printArr(long long a[], long long n) {
  for (long long i = 0; i < n; i++) cout << a[i] << " ";
  cout << '\n';
}
void printVector(std::vector<long long> v) {
  for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << '\n';
}
void printVectorPair(std::vector<pair<long long, long long>> v) {
  for (long long i = 0; i < v.size(); i++)
    cout << v[i].first << " " << v[i].second << '\n';
  ;
  cout << '\n';
}
void initialize(long long arr[], long long n) {
  for (long long i = 0; i <= n; i++) arr[i] = i;
}
long long root(long long arr[], long long i) {
  while (arr[i] != i) {
    arr[i] = arr[arr[i]];
    i = arr[i];
  }
  return i;
}
void Union(long long arr[], long long a, long long b) {
  long long root_a = root(arr, a);
  long long root_b = root(arr, b);
  arr[root_a] = root_b;
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long power_wm(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
std::vector<long long> vsum(std::vector<long long> a) {
  std::vector<long long> s(a.size());
  s[0] = a[0];
  for (long long i = 1; i <= a.size() - 1; i++) {
    s[i] = s[i - 1] + a[i];
  }
  return s;
}
void time() {}
long long sti(string s) {
  long long ans = 0;
  long long p = 1;
  for (long long i = s.size() - 1; i >= 0; i--) {
    ans = (ans + ((long long)(s[i] - '0') * p) % 1000000007) % 1000000007;
    p = (p * 10) % 1000000007;
  }
  return ans;
}
long long bfs(long long x, long long y, vector<vector<long long>> &g) {
  std::vector<long long> vis(100001, 0);
  vis[x] = 1;
  queue<long long> q;
  q.push(x);
  std::vector<long long> level(100001, (long long)1e18);
  level[x] = 0;
  while (!q.empty()) {
    long long p = q.front();
    q.pop();
    for (long long i = 0; i < g[p].size(); i++) {
      long long p2 = g[p][i];
      if (vis[p2] == 0) {
        if (x == p and y == p2) {
          continue;
        } else {
          q.push(p2);
          vis[p2] = 1;
          level[p2] = level[p] + 1;
        }
      }
    }
  }
  return level[y];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long N = 5000001;
  std::vector<long long> lp(N, 0);
  std::vector<long long> pr;
  for (long long i = 2; i <= N - 1; i++) {
    if (lp[i] == 0) {
      lp[i] = i;
      pr.push_back(i);
    }
    for (long long j = 0;
         j < (long long)pr.size() and pr[j] <= lp[i] and i * pr[j] <= N; j++)
      lp[i * pr[j]] = pr[j];
  }
  lp[1] = 1;
  std::vector<long long> count(N, 0);
  for (long long i = 2; i <= N - 1; i++) {
    long long prm = i;
    count[i]++;
    while (lp[prm] != prm) {
      count[i]++;
      prm = prm / lp[prm];
    }
  }
  count = vsum(count);
  long long t;
  cin >> t;
  while (t--) {
    long long a, b;
    cin >> a >> b;
    cout << count[a] - count[b] << '\n';
  }
  return 0;
}