#include <bits/stdc++.h>
using namespace std;
const long long int INF = 1e18;
const int N = 5e5 + 7;
const int MOD = 1e9 + 7;
constexpr int mul(int a, int b) { return 1LL * a * b % MOD; }
template <typename... T>
constexpr int mul(int a, int b, T... t) {
  return mul(mul(a, b), t...);
}
inline int read() {
  char ch = getchar();
  int x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline long long int readl() {
  char ch = getchar();
  long long int x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int A[N], ans[1010][1010];
int main() {
  int q = read();
  int MAX = 0;
  for (int i = 1; i <= q; i++) {
    int op = read(), x = read(), y = read();
    if (op == 1) {
      A[x] += y;
      for (int i = 1; i < 1010; i++) {
        ans[i][x % i] += y;
      }
    } else {
      long long int res = 0;
      if (x >= 1010) {
        for (int i = y; i < N; i += x) {
          res += A[i];
        }
      } else {
        res = ans[x][y];
      }
      cout << res << endl;
    }
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

You are given an array a consisting of 500000 integers (numbered from 1 to 500000). Initially all elements of a are zero.

You have to process two types of queries to this array:

  * 1 x y — increase a_x by y; 
  * 2 x y — compute ∑_{i ∈ R(x, y)} a_i, where R(x, y) is the set of all integers from 1 to 500000 which have remainder y modulo x. 



Can you process all the queries?

Input

The first line contains one integer q (1 ≤ q ≤ 500000) — the number of queries.

Then q lines follow, each describing a query. The i-th line contains three integers t_i, x_i and y_i (1 ≤ t_i ≤ 2). If t_i = 1, then it is a query of the first type, 1 ≤ x_i ≤ 500000, and -1000 ≤ y_i ≤ 1000. If t_i = 2, then it it a query of the second type, 1 ≤ x_i ≤ 500000, and 0 ≤ y_i < x_i.

It is guaranteed that there will be at least one query of type 2.

Output

For each query of type 2 print one integer — the answer to it.

Example

Input


5
1 3 4
2 3 0
2 4 3
1 4 -4
2 1 0


Output


4
4
0

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long int INF = 1e18;
const int N = 5e5 + 7;
const int MOD = 1e9 + 7;
constexpr int mul(int a, int b) { return 1LL * a * b % MOD; }
template <typename... T>
constexpr int mul(int a, int b, T... t) {
  return mul(mul(a, b), t...);
}
inline int read() {
  char ch = getchar();
  int x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline long long int readl() {
  char ch = getchar();
  long long int x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int A[N], ans[1010][1010];
int main() {
  int q = read();
  int MAX = 0;
  for (int i = 1; i <= q; i++) {
    int op = read(), x = read(), y = read();
    if (op == 1) {
      A[x] += y;
      for (int i = 1; i < 1010; i++) {
        ans[i][x % i] += y;
      }
    } else {
      long long int res = 0;
      if (x >= 1010) {
        for (int i = y; i < N; i += x) {
          res += A[i];
        }
      } else {
        res = ans[x][y];
      }
      cout << res << endl;
    }
  }
  return 0;
}

```