Yaroslav has n points that lie on the Ox axis. The coordinate of the first point is x1, the coordinate of the second point is x2, ..., the coordinate of the n-th point is — xn. Now Yaroslav wants to execute m queries, each of them is of one of the two following types:

  1. Move the pj-th point from position xpj to position xpj + dj. At that, it is guaranteed that after executing such query all coordinates of the points will be distinct. 
  2. Count the sum of distances between all pairs of points that lie on the segment [lj, rj] (lj ≤ rj). In other words, you should count the sum of: <image>. 



Help Yaroslav.

Input

The first line contains integer n — the number of points (1 ≤ n ≤ 105). The second line contains distinct integers x1, x2, ..., xn — the coordinates of points (|xi| ≤ 109).

The third line contains integer m — the number of queries (1 ≤ m ≤ 105). The next m lines contain the queries. The j-th line first contains integer tj (1 ≤ tj ≤ 2) — the query type. If tj = 1, then it is followed by two integers pj and dj (1 ≤ pj ≤ n, |dj| ≤ 1000). If tj = 2, then it is followed by two integers lj and rj ( - 109 ≤ lj ≤ rj ≤ 109).

It is guaranteed that at any moment all the points have distinct coordinates.

Output

For each type 2 query print the answer on a single line. Print the answers in the order, in which the queries follow in the input.

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.

Examples

Input

8
36 50 28 -75 40 -60 -95 -48
20
2 -61 29
1 5 -53
1 1 429
1 5 130
2 -101 -71
2 -69 53
1 1 404
1 5 518
2 -101 53
2 50 872
1 1 -207
2 -99 -40
1 7 -389
1 6 -171
1 2 464
1 7 -707
1 1 -730
1 1 560
2 635 644
1 7 -677


Output

176
20
406
1046
1638
156
0
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 9;
long long arrr[N], pl[N];
pair<long long, long long> arr[N];
long long t[N], sum[N];
int n;
void update(int i, int j, int v = 1, int b = 0, int e = n - 1) {
  if (i > e || b > j) return;
  if (b == e) {
    t[v] = 0;
    sum[v] = arr[b].first;
    return;
  }
  long long mid = b + e >> 1, l = v << 1, r = l | 1, sz1 = mid - b + 1,
            sz2 = e - mid;
  update(i, j, l, b, mid);
  update(i, j, r, mid + 1, e);
  t[v] = t[l] + t[r] + sz2 * -sum[l] + sz1 * sum[r];
  sum[v] = sum[l] + sum[r];
  return;
}
long long query(int i, int j, int v = 1, int b = 0, int e = n - 1) {
  if (i > e || b > j) return 0;
  if (i <= b && e <= j) {
    return t[v] + ((long long)b - i) * sum[v] - ((long long)j - e) * sum[v];
  }
  int mid = b + e >> 1, l = v << 1, r = l | 1;
  return query(i, j, l, b, mid) + query(i, j, r, mid + 1, e);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> arr[i].first, arr[i].second = i;
  sort(arr, arr + n);
  for (int i = 0; i < n; i++) pl[arr[i].second] = i;
  update(0, n - 1);
  int m;
  cin >> m;
  while (m--) {
    int c;
    cin >> c;
    --c;
    if (c) {
      long long lf, rg;
      cin >> lf >> rg;
      int l = lower_bound(arr, arr + n, make_pair(lf, (long long)-1)) - arr;
      int r = lower_bound(arr, arr + n, make_pair(rg + 1, (long long)-1)) - arr;
      --r;
      if (r < l)
        cout << 0 << '\n';
      else
        cout << query(l, r) << '\n';
    } else {
      int l = 0, r = 0;
      int v, d;
      cin >> v >> d;
      --v;
      int id = pl[v];
      arr[id].first += d;
      l = id;
      r = id;
      if (id && arr[id].first < arr[id - 1].first) {
        while (l && arr[l].first < arr[l - 1].first)
          swap(arr[l], arr[l - 1]), l, pl[arr[l].second]++,
              pl[arr[l - 1].second]--, l--;
      } else {
        while (r != n - 1 && arr[r].first > arr[r + 1].first)
          swap(arr[r], arr[r + 1]), r, pl[arr[r + 1].second]++,
              pl[arr[r].second]--, r++;
      }
      update(l, r);
    }
  }
}
