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 MAX = int(1e5) + 10;
struct node {
  int cc;
  long long sum1, sum2;
} T[MAX << 3];
int N, M;
int a[MAX], cur[MAX];
int typ[MAX], x[MAX], y[MAX];
vector<int> X;
int getID(int x) { return lower_bound(X.begin(), X.end(), x) - X.begin(); }
node merge(node &L, node &R) {
  node ret;
  ret.cc = L.cc + R.cc;
  ret.sum1 = L.sum1 + R.sum1;
  ret.sum2 = L.sum2 + R.sum2 + L.cc * R.sum1;
  return ret;
}
void update(int nod, int lo, int hi, int p, int v, int c) {
  if (p < lo || hi < p) return;
  if (lo == p && hi == p) {
    T[nod].cc += c;
    T[nod].sum1 += v;
    T[nod].sum2 = T[nod].cc * T[nod].sum1;
    return;
  }
  int mit = (lo + hi) >> 1, I = nod << 1, D = I + 1;
  update(I, lo, mit, p, v, c);
  update(D, mit + 1, hi, p, v, c);
  T[nod] = merge(T[I], T[D]);
}
node query(int nod, int lo, int hi, int a, int b) {
  if (b < lo || hi < a) return (node){0, 0, 0};
  if (a <= lo && hi <= b) return T[nod];
  int mit = (lo + hi) >> 1, I = nod << 1, D = I + 1;
  node L = query(I, lo, mit, a, b);
  node R = query(D, mit + 1, hi, a, b);
  return merge(L, R);
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    scanf("%d", a + i);
    cur[i] = a[i];
    X.push_back(a[i]);
  }
  int Q;
  scanf("%d", &Q);
  for (int i = 0; i < Q; i++) {
    scanf("%d%d%d", typ + i, x + i, y + i);
    if (typ[i] == 1) {
      x[i]--;
      cur[x[i]] += y[i];
      X.push_back(cur[x[i]]);
    }
  }
  sort(X.begin(), X.end());
  X.erase(unique(X.begin(), X.end()), X.end());
  M = (int)X.size();
  for (int i = 0; i < N; i++) {
    update(1, 0, M - 1, getID(a[i]), a[i], 1);
    cur[i] = a[i];
  }
  for (int i = 0; i < Q; i++) {
    if (typ[i] == 1) {
      update(1, 0, M - 1, getID(cur[x[i]]), -cur[x[i]], -1);
      cur[x[i]] += y[i];
      update(1, 0, M - 1, getID(cur[x[i]]), cur[x[i]], 1);
    } else {
      int L = lower_bound(X.begin(), X.end(), x[i]) - X.begin();
      int R = upper_bound(X.begin(), X.end(), y[i]) - X.begin();
      long long ans = 0;
      if (L < R) {
        node seg = query(1, 0, M - 1, L, R - 1);
        ans = 2 * seg.sum2 - (seg.cc + 1) * seg.sum1;
      }
      printf("%I64d\n", ans);
    }
  }
  return 0;
}
