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;
struct T {
  long long sum;
  int len;
  long long sol;
};
T unite(T a, T b) {
  long long sum = a.sum + b.sum;
  int len = a.len + b.len;
  long long sol = a.sol + b.sol + b.sum * a.len - a.sum * b.len;
  return {sum, len, sol};
}
const int N = (int)1e5 + 7;
int n;
int a[N];
int grow[N];
int m;
map<int, int> trans;
int tp[N];
int x[N];
int y[N];
int now;
T aint[4 * N * 3];
void del(int v, int tl, int tr, int pos) {
  if (tr < pos || pos < tl) {
    return;
  }
  if (tl == tr) {
    aint[v] = {0, 0, 0};
  } else {
    int tm = (tl + tr) / 2;
    del(2 * v, tl, tm, pos);
    del(2 * v + 1, tm + 1, tr, pos);
    aint[v] = unite(aint[2 * v], aint[2 * v + 1]);
  }
}
void make(int v, int tl, int tr, int pos, long long x) {
  if (tr < pos || pos < tl) {
    return;
  }
  if (tl == tr) {
    aint[v] = {x, 1, 0};
  } else {
    int tm = (tl + tr) / 2;
    make(2 * v, tl, tm, pos, x);
    make(2 * v + 1, tm + 1, tr, pos, x);
    aint[v] = unite(aint[2 * v], aint[2 * v + 1]);
  }
}
T get(int v, int tl, int tr, int l, int r) {
  if (tr < l || r < tl) {
    return {0, 0, 0};
  }
  if (l <= tl && tr <= r) {
    return aint[v];
  } else {
    int tm = (tl + tr) / 2;
    return unite(get(2 * v, tl, tm, l, r), get(2 * v + 1, tm + 1, tr, l, r));
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    trans[a[i]] = 0;
  }
  cin >> m;
  for (int i = 1; i <= m; i++) {
    cin >> tp[i] >> x[i] >> y[i];
    if (tp[i] == 1) {
      grow[x[i]] += y[i];
      trans[a[x[i]] + grow[x[i]]] = 0;
    } else {
      trans[x[i]] = 0;
      trans[y[i]] = 0;
    }
  }
  for (auto &it : trans) {
    it.second = ++now;
  }
  for (int i = 1; i <= n; i++) {
    make(1, 1, now, trans[a[i]], a[i]);
  }
  for (int i = 1; i <= m; i++) {
    if (tp[i] == 1) {
      del(1, 1, now, trans[a[x[i]]]);
      a[x[i]] += y[i];
      make(1, 1, now, trans[a[x[i]]], a[x[i]]);
    } else {
      auto it = get(1, 1, now, trans[x[i]], trans[y[i]]);
      cout << it.sol << "\n";
    }
  }
}
