#include <bits/stdc++.h>
using namespace std;
template <class T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
template <class T>
void inp(T& x) {
  cin >> x;
}
template <class H, class... T>
void inp(H& h, T&... t) {
  inp(h);
  inp(t...);
}
template <class A>
void inp(vector<A>& x) {
  for (auto& a : x) inp(a);
}
void deb_out() { cerr << endl; }
template <typename Head, typename... Tail>
void deb_out(Head H, Tail... T) {
  cerr << ' ' << H;
  deb_out(T...);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int mod = 1000000007;
const char nl = '\n';
void solve() {
  long long l, r;
  inp(l, r);
  if (r % 2 == 0) {
    if (l > r / 2 + 1)
      cout << r % l << nl;
    else
      cout << r % (r / 2 + 1) << nl;
  } else {
    if (l > (r + 1) / 2)
      cout << r % l << nl;
    else
      cout << r % ((r + 1) / 2) << nl;
  }
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
  int t = 1;
  inp(t);
  while (t--) solve();
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

You are given two integers l and r, l≤ r. Find the largest possible value of a mod b over all pairs (a, b) of integers for which r≥ a ≥ b ≥ l.

As a reminder, a mod b is a remainder we get when dividing a by b. For example, 26 mod 8 = 2.

Input

Each test contains multiple test cases.

The first line contains one positive integer t (1≤ t≤ 10^4), denoting the number of test cases. Description of the test cases follows.

The only line of each test case contains two integers l, r (1≤ l ≤ r ≤ 10^9).

Output

For every test case, output the largest possible value of a mod b over all pairs (a, b) of integers for which r≥ a ≥ b ≥ l.

Example

Input


4
1 1
999999999 1000000000
8 26
1 999999999


Output


0
1
12
499999999

Note

In the first test case, the only allowed pair is (a, b) = (1, 1), for which a mod b = 1 mod 1 = 0.

In the second test case, the optimal choice is pair (a, b) = (1000000000, 999999999), for which a mod b = 1.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
template <class T>
void inp(T& x) {
  cin >> x;
}
template <class H, class... T>
void inp(H& h, T&... t) {
  inp(h);
  inp(t...);
}
template <class A>
void inp(vector<A>& x) {
  for (auto& a : x) inp(a);
}
void deb_out() { cerr << endl; }
template <typename Head, typename... Tail>
void deb_out(Head H, Tail... T) {
  cerr << ' ' << H;
  deb_out(T...);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int mod = 1000000007;
const char nl = '\n';
void solve() {
  long long l, r;
  inp(l, r);
  if (r % 2 == 0) {
    if (l > r / 2 + 1)
      cout << r % l << nl;
    else
      cout << r % (r / 2 + 1) << nl;
  } else {
    if (l > (r + 1) / 2)
      cout << r % l << nl;
    else
      cout << r % ((r + 1) / 2) << nl;
  }
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
  int t = 1;
  inp(t);
  while (t--) solve();
  return 0;
}

```