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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimization("unroll-loops")
#pragma GCC optimization("O3")
using namespace std;
const long long N = 1e5 + 10, LOG = 26, M = 2e9 + 10, Mod = 1e9 + 9;
struct node {
  long long sum, cnt, ans;
  node* lc;
  node* rc;
  node() : lc(NULL), rc(NULL), ans(0), sum(0), cnt(0) {}
  node(long long ans, long long sum, long long cnt)
      : lc(NULL), rc(NULL), ans(ans), sum(sum), cnt(cnt) {}
};
void merge(node* it) {
  it->sum = it->lc->sum + it->rc->sum;
  it->ans = it->lc->ans + it->rc->ans + it->rc->sum * it->lc->cnt -
            it->lc->sum * it->rc->cnt;
  it->cnt = it->lc->cnt + it->rc->cnt;
}
void make(node* it) {
  if (it->lc == NULL) {
    node* x = new node;
    it->lc = x;
  }
  if (it->rc == NULL) {
    node* x = new node;
    it->rc = x;
  }
}
void ins(long long p, node* it, long long l = -M, long long r = M) {
  if (r - l < 2) {
    it->ans = 0;
    it->cnt = 1;
    it->sum = p;
    return;
  }
  make(it);
  long long mid = (l + r) / 2;
  if (p < mid)
    ins(p, it->lc, l, mid);
  else
    ins(p, it->rc, mid, r);
  merge(it);
}
void del(long long p, node* it, long long l = -M, long long r = M) {
  if (r - l < 2) {
    it->ans = it->cnt = it->sum = 0;
    return;
  }
  make(it);
  long long mid = (l + r) / 2;
  if (p < mid)
    del(p, it->lc, l, mid);
  else
    del(p, it->rc, mid, r);
  merge(it);
}
node que(long long s, long long e, node* it, long long l = -M,
         long long r = M) {
  if (s <= l && r <= e) return *it;
  if (r <= s || e <= l) return node(0, 0, 0);
  make(it);
  long long mid = (l + r) / 2;
  node a = que(s, e, it->lc, l, mid);
  node b = que(s, e, it->rc, mid, r);
  return node(a.ans + b.ans + b.sum * a.cnt - a.sum * b.cnt, a.sum + b.sum,
              a.cnt + b.cnt);
}
long long point[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  node* root = new node;
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> point[i];
    ins(point[i], root);
  }
  int q;
  cin >> q;
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int p, d;
      cin >> p >> d;
      p--;
      del(point[p], root);
      point[p] += d;
      ins(point[p], root);
    } else {
      int l, r;
      cin >> l >> r;
      r++;
      cout << que(l, r, root).ans << '\n';
    }
  }
}
