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>
int inp() {
  char c = getchar();
  int neg = 1;
  while (c < '0' || c > '9') {
    if (c == '-') neg = -1;
    c = getchar();
  }
  int sum = 0;
  while (c >= '0' && c <= '9') {
    sum = sum * 10 + c - '0';
    c = getchar();
  }
  return sum * neg;
}
int root = 0;
int x[100010];
struct fhq_Treap {
  struct T {
    int lc;
    int rc;
    long long sum;
    long long sum2;
    int size;
    int v;
    int w;
  } t[200010];
  int cnt;
  void push_up(int cur) {
    t[cur].size = t[t[cur].lc].size + t[t[cur].rc].size + 1;
    t[cur].sum2 = t[t[cur].lc].sum2 + t[t[cur].rc].sum2 + t[cur].v;
    t[cur].sum = t[t[cur].lc].sum + t[t[cur].rc].sum +
                 (t[t[cur].rc].sum2 + t[cur].v) * (t[t[cur].lc].size + 1);
  }
  int merge(int a, int b) {
    if (!a || !b) return a + b;
    if (t[a].w < t[b].w) {
      t[a].rc = merge(t[a].rc, b);
      push_up(a);
      return a;
    } else {
      t[b].lc = merge(a, t[b].lc);
      push_up(b);
      return b;
    }
  }
  void split(int cur, int &a, int &b, int k) {
    int cc = 1e8;
    if (!cur) {
      a = b = 0;
      return;
    }
    if (t[cur].v <= k) {
      a = cur;
      split(t[cur].rc, t[a].rc, b, k);
      push_up(a);
    } else {
      b = cur;
      split(t[cur].lc, a, t[b].lc, k);
      push_up(cur);
    }
  }
  void print(int cur) {
    if (cur == 0) return;
    print(t[cur].lc);
    print(t[cur].rc);
  }
  long long query(int l, int r) {
    int lb, mb, rb;
    split(root, lb, rb, l - 1);
    split(rb, mb, rb, r);
    long long ans = t[mb].sum * 2 - (t[mb].sum2 * (t[mb].size + 1));
    root = merge(merge(lb, mb), rb);
    return ans;
  }
  void erase(int x) {
    int l, m, r;
    split(root, l, r, x - 1);
    split(r, m, r, x);
    root = merge(l, r);
  }
  void insert(int x) {
    t[++cnt] = (T){0, 0, x, x, 1, x, rand()};
    int l, r;
    split(root, l, r, x);
    root = merge(merge(l, cnt), r);
  }
} t;
int main() {
  int n = inp();
  for (int i = 1; i <= n; i++) {
    x[i] = inp();
    t.insert(x[i]);
  }
  int m = inp();
  while (m--) {
    int op = inp();
    if (op == 1) {
      int p = inp();
      int d = inp();
      t.erase(x[p]);
      x[p] += d;
      t.insert(x[p]);
    } else {
      int l = inp();
      int r = inp();
      std::cout << t.query(l, r) << std::endl;
    }
  }
}
