观察区间 $[a^2, (a+1)^2)$ 中有多少个数字满足条件，满足条件的数字一定有因数 $a$，满足形式：$a, a(a+1), a(a+2), \ldots$，又注意到 $(a+1)^2 - 1 = a(a+2)$，因此每个区间只有三个数字满足条件。剩下的就是区间处理了，先计算出完整的区间个数，然后考虑两端的不完整区间中满足条件的数字。数字范围是 $10^{18}$，求平方根使用二分，二分求中点可能会溢出，把右边界设置成最大可能的平方根即可。

// Date: Tue Nov 21 19:23:45 2023

#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>

#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

using namespace std;

const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
const int dir[8][2] = {
    {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;

const ull Pr = 131;

#define LN ListNode
#define LNP ListNode *
#define TN TreeNode
#define TNP TreeNode *

#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])

#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nonempty(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())

#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z)                                                        \
  cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#else
#define debug1
#define debug2
#define debug3
#endif

#ifdef _DEBUG

struct ListNode {
  int val;
  ListNode *next;
  ListNode() : val(0), next(nullptr) {}
  ListNode(int val) : val(val), next(nullptr) {}
  ListNode(int val, ListNode *next) : val(val), next(next) {}
};

struct TreeNode {
  int val;
  TreeNode *left;
  TreeNode *right;
  TreeNode() : val(0), left(nullptr), right(nullptr) {}
  TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
  TreeNode(int x, TreeNode *left, TreeNode *right)
      : val(x), left(left), right(right) {}
};

#endif

ull a, b;
int t;

bool check(ull x) { return x >= a && x <= b; }

ull sqrt1(ull x) {
  if (x == 1)
    return 1;

  ull l = 1, r = 1e9 + 10, m;
  while (l < r) {
    m = (l + r + 1) / 2;
    if (m * m <= x)
      l = m;
    else
      r = m - 1;
  }

  return l;
}

int main(void) {
#ifdef _DEBUG
  freopen("1737b.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  int t;
  cin >> t;
  while (t--) {
    cin >> a >> b;

    ull a1 = sqrt1(a), b1 = sqrt1(b), ans = 0;

    if (a1 == b1) {
      ans = 0;
      For(i, 0, 3) {
        if (check(a1 * (a1 + i)))
          ans++;
      }
      cout << ans << '\n';
      continue;
    }

    ans = (b1 - a1 - 1) * 3;
    For(i, 0, 3) {
      if (check(a1 * (a1 + i)))
        ans++;
      if (check(b1 * (b1 + i)))
        ans++;
    }

    cout << ans << '\n';
  }

  return 0;
}