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("O2,Ofast,inline,unroll-all-loops,-ffast-math")
#pragma GCC target("popcnt")
using namespace std;
struct node {
  int l, r, cnt;
  long long sum, ans;
  node(int _cnt = 0, long long _sum = 0, long long _ans = 0) {
    cnt = _cnt, sum = _sum, ans = _ans, l = r = 0;
  }
  node operator+(const node &rhs) const {
    return node(cnt + rhs.cnt, sum + rhs.sum,
                ans + rhs.ans + rhs.sum * cnt - sum * rhs.cnt);
  }
} t[100010 * 79];
int a[100010], n, q, rt, cnt = 0;
const int INF = 2e9 + 10;
template <class T>
void read(T &x) {
  char ch = x = 0;
  bool fl = false;
  while (!isdigit(ch)) fl |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  x = fl ? -x : x;
}
void pushup(int x) {
  int l = t[x].l, r = t[x].r;
  t[x] = t[t[x].l] + t[t[x].r], t[x].l = l, t[x].r = r;
}
void insert(int &x, int pos, int l, int r, int type) {
  if (!x) x = ++cnt;
  if (l == r) return t[x] = node(type, type * l, 0), void();
  int mid = (long long)l + r >> 1;
  if (pos <= mid)
    insert(t[x].l, pos, l, mid, type);
  else
    insert(t[x].r, pos, mid + 1, r, type);
  pushup(x);
}
node query(int x, int ql, int qr, int l, int r) {
  if (!x) return node();
  if (ql <= l && r <= qr) return t[x];
  int mid = (long long)l + r >> 1;
  node ret = node();
  if (ql <= mid) ret = ret + query(t[x].l, ql, qr, l, mid);
  if (qr >= mid + 1) ret = ret + query(t[x].r, ql, qr, mid + 1, r);
  return ret;
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) {
    read(a[i]), insert(rt, a[i], -INF, INF, 1);
  }
  read(q);
  for (int i = 1, op, x, y; i <= q; i++) {
    read(op), read(x), read(y);
    if (op == 1) {
      insert(rt, a[x], -INF, INF, 0);
      insert(rt, a[x] += y, -INF, INF, 1);
    } else {
      printf("%lld\n", query(rt, x, y, -INF, INF).ans);
    }
  }
  return 0;
}
