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;
int n, m, x[100005], a[100005 * 3], aa[100005 * 3], tot2, xx[100005], tot;
struct Tnode {
  long long sum, cnt;
  long long tot;
} tree[100005 * 12];
struct node {
  int l, r, num;
} query[100005];
void init() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x[i]);
    xx[i] = x[i];
    aa[++tot2] = x[i];
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &query[i].num, &query[i].l, &query[i].r);
    if (query[i].num == 1) {
      xx[query[i].l] += query[i].r;
      aa[++tot2] = xx[query[i].l];
    } else {
      aa[++tot2] = query[i].l;
      aa[++tot2] = query[i].r;
    }
  }
}
void ins(int step, int l, int r, int x, int y) {
  if (l == r) {
    if (y) {
      tree[step].cnt = a[x];
      tree[step].tot = 1;
    } else
      tree[step].cnt = tree[step].tot = 0;
    return;
  }
  int mid = (l + r) >> 1;
  if (x <= mid)
    ins(step << 1, l, mid, x, y);
  else
    ins(step << 1 | 1, mid + 1, r, x, y);
  tree[step].tot = tree[step << 1].tot + tree[step << 1 | 1].tot;
  tree[step].cnt = tree[step << 1].cnt + tree[step << 1 | 1].cnt;
  tree[step].sum =
      tree[step << 1].sum + tree[step << 1].tot * tree[step << 1 | 1].cnt -
      tree[step << 1 | 1].tot * tree[step << 1].cnt + tree[step << 1 | 1].sum;
}
Tnode ask(int step, int l, int r, int ll, int rr) {
  Tnode t1, t2, t;
  if (ll <= l && rr >= r) return tree[step];
  int mid = (l + r) >> 1;
  if (ll <= mid)
    t1 = ask(step << 1, l, mid, ll, rr);
  else {
    t1.sum = t1.cnt = t1.tot = 0;
  }
  if (rr > mid)
    t2 = ask(step << 1 | 1, mid + 1, r, ll, rr);
  else
    t2.sum = t2.cnt = t2.tot = 0;
  t.tot = t1.tot + t2.tot;
  t.cnt = t1.cnt + t2.cnt;
  t.sum = t1.sum + t1.tot * t2.cnt - t2.tot * t1.cnt + t2.sum;
  return t;
}
void work() {
  int i, j, k;
  Tnode t;
  sort(aa + 1, aa + tot2 + 1);
  a[1] = aa[1];
  tot = 1;
  for (i = 2; i <= tot2; i++)
    if (aa[i] != aa[i - 1]) a[++tot] = aa[i];
  for (i = 1; i <= n; i++)
    ins(1, 1, tot, lower_bound(a + 1, a + tot + 1, x[i]) - a, 1);
  for (i = 1; i <= m; i++) {
    if (query[i].num == 1) {
      ins(1, 1, tot, lower_bound(a + 1, a + tot + 1, x[query[i].l]) - a, 0);
      x[query[i].l] += query[i].r;
      ins(1, 1, tot, lower_bound(a + 1, a + tot + 1, x[query[i].l]) - a, 1);
    } else {
      t = ask(1, 1, tot, lower_bound(a + 1, a + tot + 1, query[i].l) - a,
              lower_bound(a + 1, a + tot + 1, query[i].r) - a);
      printf("%I64d\n", t.sum);
    }
  }
}
int main() {
  init();
  work();
  return 0;
}
