#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> PII;
int dir[8][2] = {-1, 0, 1, 0, 0, -1, 0, 1, -1, -1, -1, 1, 1 - 1, 1, 1};
int sgn(double x) {
  if (fabs(x) < 1e-8) return 0;
  if (x < 0)
    return -1;
  else
    return 1;
}
long long qpow(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % m;
    a = a * a % m, b >>= 1;
  }
  return ans;
}
long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a;
    a = a * a, b >>= 1;
  }
  return ans;
}
long long qmul(long long a, long long b, long long m) {
  long long res = 0;
  while (b > 0) {
    if (b & 1) res = (res + a) % m;
    a = (a + a) % m;
    b >>= 1;
  }
  return res;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long inv(long long t, long long p) {
  return t == 1 ? 1 : (p - p / t) * inv(p % t, p) % p;
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long r = exgcd(b, a % b, y, x);
  y -= (a / b) * x;
  return r;
}
long long inv1(long long a, long long p) {
  long long d, x, y;
  d = exgcd(a, p, x, y);
  return d == 1 ? (x % p + p) % p : -1;
}
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
template <typename S, typename T>
inline bool Min(S &a, const T &b) {
  return a > b ? a = b, true : false;
}
template <typename S, typename T>
inline bool Max(S &a, const T &b) {
  return a < b ? a = b, true : false;
}
template <typename T>
inline bool IsPri(T x) {
  if (x < 2) return false;
  for (T i = 2; i * i <= x; ++i)
    if (x % i == 0) return false;
  return true;
}
template <typename T>
inline int BitCnt(T x) {
  int cnt = 0;
  while (x) ++cnt, x &= x - 1;
  return cnt;
}
template <typename T>
inline int BitNum(T x) {
  int cnt = 0;
  while (x) ++cnt, x >>= 1;
  return cnt;
}
template <class T>
void _sf(T &x) {
  cin >> x;
}
void _sf(int &x) { scanf("%d", &x); }
void _sf(long long &x) { scanf("%lld", &x); }
void _sf(double &x) { scanf("%lf", &x); }
void _sf(char &x) { scanf(" %c", &x); }
void _sf(char *x) { scanf("%s", x); }
void sf() {}
template <class T, class... U>
void sf(T &head, U &...tail) {
  _sf(head);
  sf(tail...);
}
template <class T>
void _pf(const T &x) {
  cout << x;
}
void _pf(const int &x) { printf("%d", x); }
void _pf(const long long &x) { printf("%lld", x); }
void _pf(const double &x) { printf("%.16f", x); }
void _pf(const char &x) { putchar(x); }
void _pf(const char *x) { printf("%s", x); }
template <class T, class U>
void _pf(const pair<T, U> &x) {
  pf(x.F);
  putchar(' ');
  pf(x.S);
}
template <class T>
void _pf(const vector<T> &x) {
  for (auto i = x.begin(); i != x.end(); _pf(*i++))
    if (i != x.cbegin()) putchar(' ');
}
void pf() {}
template <class T, class... U>
void pf(const T &head, const U &...tail) {
  _pf(head);
  putchar(sizeof...(tail) ? ' ' : '\n');
  pf(tail...);
}
template <typename T>
inline void read(T &x) {
  x = 0;
  static int p;
  p = 1;
  static char c;
  c = getchar();
  while (!isdigit(c)) {
    if (c == '-') p = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c - 48);
    c = getchar();
  }
  x *= p;
}
const bool IOS = false;
const double PI = acos(-1);
const double eps = 1e-8;
const long long linf = 0x3f3f3f3f3f3f3f3f;
const long long INV2 = 500000004;
const int mod = 998244353;
const int inf = 0x3f3f3f3f;
const int maxn = 1e6 + 10;
void Main();
int main() {
  if (IOS) std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  Main();
  0;
  return 0;
}
void Main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    long long n, r;
    sf(n, r);
    long long ans = 0;
    if (n <= r) {
      ans = n * (n - 1) / 2 + 1;
    } else
      ans = (1 + r) * r / 2;
    pf(ans);
  }
}

### Prompt

Create a solution in CPP for the following problem:

A competitive eater, Alice is scheduling some practices for an eating contest on a magical calendar. The calendar is unusual because a week contains not necessarily 7 days!

In detail, she can choose any integer k which satisfies 1 ≤ k ≤ r, and set k days as the number of days in a week.

Alice is going to paint some n consecutive days on this calendar. On this calendar, dates are written from the left cell to the right cell in a week. If a date reaches the last day of a week, the next day's cell is the leftmost cell in the next (under) row.

She wants to make all of the painted cells to be connected by side. It means, that for any two painted cells there should exist at least one sequence of painted cells, started in one of these cells, and ended in another, such that any two consecutive cells in this sequence are connected by side.

Alice is considering the shape of the painted cells. Two shapes are the same if there exists a way to make them exactly overlapped using only parallel moves, parallel to the calendar's sides.

For example, in the picture, a week has 4 days and Alice paints 5 consecutive days. [1] and [2] are different shapes, but [1] and [3] are equal shapes.

<image>

Alice wants to know how many possible shapes exists if she set how many days a week has and choose consecutive n days and paints them in calendar started in one of the days of the week. As was said before, she considers only shapes, there all cells are connected by side.

Input

The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.

For each test case, the only line contains two integers n, r (1 ≤ n ≤ 10^9, 1 ≤ r ≤ 10^9).

Output

For each test case, print a single integer — the answer to the problem.

Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language.

Example

Input


5
3 4
3 2
3 1
13 7
1010000 9999999


Output


4
3
1
28
510049495001

Note

In the first test case, Alice can set 1,2,3 or 4 days as the number of days in a week.

There are 6 possible paintings shown in the picture, but there are only 4 different shapes. So, the answer is 4. Notice that the last example in the picture is an invalid painting because all cells are not connected by sides.

<image>

In the last test case, be careful with the overflow issue, described in the output format.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> PII;
int dir[8][2] = {-1, 0, 1, 0, 0, -1, 0, 1, -1, -1, -1, 1, 1 - 1, 1, 1};
int sgn(double x) {
  if (fabs(x) < 1e-8) return 0;
  if (x < 0)
    return -1;
  else
    return 1;
}
long long qpow(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % m;
    a = a * a % m, b >>= 1;
  }
  return ans;
}
long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a;
    a = a * a, b >>= 1;
  }
  return ans;
}
long long qmul(long long a, long long b, long long m) {
  long long res = 0;
  while (b > 0) {
    if (b & 1) res = (res + a) % m;
    a = (a + a) % m;
    b >>= 1;
  }
  return res;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long inv(long long t, long long p) {
  return t == 1 ? 1 : (p - p / t) * inv(p % t, p) % p;
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long r = exgcd(b, a % b, y, x);
  y -= (a / b) * x;
  return r;
}
long long inv1(long long a, long long p) {
  long long d, x, y;
  d = exgcd(a, p, x, y);
  return d == 1 ? (x % p + p) % p : -1;
}
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
template <typename S, typename T>
inline bool Min(S &a, const T &b) {
  return a > b ? a = b, true : false;
}
template <typename S, typename T>
inline bool Max(S &a, const T &b) {
  return a < b ? a = b, true : false;
}
template <typename T>
inline bool IsPri(T x) {
  if (x < 2) return false;
  for (T i = 2; i * i <= x; ++i)
    if (x % i == 0) return false;
  return true;
}
template <typename T>
inline int BitCnt(T x) {
  int cnt = 0;
  while (x) ++cnt, x &= x - 1;
  return cnt;
}
template <typename T>
inline int BitNum(T x) {
  int cnt = 0;
  while (x) ++cnt, x >>= 1;
  return cnt;
}
template <class T>
void _sf(T &x) {
  cin >> x;
}
void _sf(int &x) { scanf("%d", &x); }
void _sf(long long &x) { scanf("%lld", &x); }
void _sf(double &x) { scanf("%lf", &x); }
void _sf(char &x) { scanf(" %c", &x); }
void _sf(char *x) { scanf("%s", x); }
void sf() {}
template <class T, class... U>
void sf(T &head, U &...tail) {
  _sf(head);
  sf(tail...);
}
template <class T>
void _pf(const T &x) {
  cout << x;
}
void _pf(const int &x) { printf("%d", x); }
void _pf(const long long &x) { printf("%lld", x); }
void _pf(const double &x) { printf("%.16f", x); }
void _pf(const char &x) { putchar(x); }
void _pf(const char *x) { printf("%s", x); }
template <class T, class U>
void _pf(const pair<T, U> &x) {
  pf(x.F);
  putchar(' ');
  pf(x.S);
}
template <class T>
void _pf(const vector<T> &x) {
  for (auto i = x.begin(); i != x.end(); _pf(*i++))
    if (i != x.cbegin()) putchar(' ');
}
void pf() {}
template <class T, class... U>
void pf(const T &head, const U &...tail) {
  _pf(head);
  putchar(sizeof...(tail) ? ' ' : '\n');
  pf(tail...);
}
template <typename T>
inline void read(T &x) {
  x = 0;
  static int p;
  p = 1;
  static char c;
  c = getchar();
  while (!isdigit(c)) {
    if (c == '-') p = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c - 48);
    c = getchar();
  }
  x *= p;
}
const bool IOS = false;
const double PI = acos(-1);
const double eps = 1e-8;
const long long linf = 0x3f3f3f3f3f3f3f3f;
const long long INV2 = 500000004;
const int mod = 998244353;
const int inf = 0x3f3f3f3f;
const int maxn = 1e6 + 10;
void Main();
int main() {
  if (IOS) std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  Main();
  0;
  return 0;
}
void Main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    long long n, r;
    sf(n, r);
    long long ans = 0;
    if (n <= r) {
      ans = n * (n - 1) / 2 + 1;
    } else
      ans = (1 + r) * r / 2;
    pf(ans);
  }
}

```