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 maxn = 100000;
int n, m, N;
int a[maxn + 1], A[maxn + 1], b[maxn << 1 | 1];
struct Qs {
  int t, a, b;
} Q[maxn + 1];
map<int, int> Map;
bool c[maxn << 1 | 1];
struct mes {
  int cnt;
  long long sum, ans;
  mes operator+(const mes &Other) const {
    mes A;
    A.cnt = cnt + Other.cnt;
    A.sum = sum + Other.sum;
    A.ans = ans + Other.ans + Other.sum * cnt - sum * Other.cnt;
    return A;
  }
};
struct node {
  int l, r;
  mes imt;
} p[maxn << 3 | 1];
void Build(int k, int l, int r) {
  p[k].l = l;
  p[k].r = r;
  if (l == r) {
    if (c[l])
      p[k].imt = (mes){1, b[l], 0};
    else
      p[k].imt = (mes){0, 0, 0};
    return;
  }
  Build(k << 1, l, (l + r) >> 1);
  Build(k << 1 | 1, ((l + r) >> 1) + 1, r);
  p[k].imt = p[k << 1].imt + p[k << 1 | 1].imt;
  return;
}
void Insert(int k, int x) {
  if (p[k].l == x && x == p[k].r) {
    p[k].imt = (mes){1, b[x], 0};
    return;
  }
  if (x <= p[k << 1].r)
    Insert(k << 1, x);
  else
    Insert(k << 1 | 1, x);
  p[k].imt = p[k << 1].imt + p[k << 1 | 1].imt;
  return;
}
void Del(int k, int x) {
  if (p[k].l == x && x == p[k].r) {
    p[k].imt = (mes){0, 0, 0};
    return;
  }
  if (x <= p[k << 1].r)
    Del(k << 1, x);
  else
    Del(k << 1 | 1, x);
  p[k].imt = p[k << 1].imt + p[k << 1 | 1].imt;
  return;
}
mes Query(int k, int l, int r) {
  if (l <= p[k].l && p[k].r <= r) return p[k].imt;
  mes A = (mes){0, 0, 0};
  if (l <= p[k << 1].r) A = Query(k << 1, l, r);
  if (p[k << 1 | 1].l <= r) A = A + Query(k << 1 | 1, l, r);
  return A;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    b[i] = A[i] = a[i];
  }
  N = n;
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d%d", &Q[i].t, &Q[i].a, &Q[i].b);
    if (Q[i].t == 1) b[++N] = (A[Q[i].a] += Q[i].b);
  }
  sort(b + 1, b + N + 1);
  N = unique(b + 1, b + N + 1) - b - 1;
  for (int i = 1; i <= N; ++i) Map[b[i]] = i;
  for (int i = 1; i <= n; ++i) c[Map[a[i]]] = 1;
  Build(1, 1, N);
  for (int i = 1; i <= m; ++i)
    if (Q[i].t == 1) {
      Del(1, Map[a[Q[i].a]]);
      Insert(1, Map[a[Q[i].a] += Q[i].b]);
    } else {
      if (Q[i].b < b[1] || b[N] < Q[i].a) {
        printf("0\n");
        continue;
      }
      int l = lower_bound(b + 1, b + N + 1, Q[i].a) - b,
          r = upper_bound(b + 1, b + N + 1, Q[i].b) - b - 1;
      printf("%lld\n", Query(1, l, r).ans);
    }
  return 0;
}
