Fedya and Sasha are friends, that's why Sasha knows everything about Fedya.

Fedya keeps his patience in an infinitely large bowl. But, unlike the bowl, Fedya's patience isn't infinite, that is why let v be the number of liters of Fedya's patience, and, as soon as v becomes equal to 0, the bowl will burst immediately. There is one tap in the bowl which pumps s liters of patience per second. Notice that s can be negative, in that case, the tap pumps out the patience. Sasha can do different things, so he is able to change the tap's speed. All actions that Sasha does can be represented as q queries. There are three types of queries:

  1. "1 t s" — add a new event, means that starting from the t-th second the tap's speed will be equal to s. 
  2. "2 t" — delete the event which happens at the t-th second. It is guaranteed that such event exists. 
  3. "3 l r v" — Sasha wonders: if you take all the events for which l ≤ t ≤ r and simulate changes of Fedya's patience from the very beginning of the l-th second till the very beginning of the r-th second inclusive (the initial volume of patience, at the beginning of the l-th second, equals to v liters) then when will be the moment when the bowl will burst. If that does not happen, then the answer will be -1. 



Since Sasha does not want to check what will happen when Fedya's patience ends, and he has already come up with the queries, he is asking you to help him and find the answer for each query of the 3-rd type.

It is guaranteed that at any moment of time, there won't be two events which happen at the same second.

Input

The first line contans one integer q (1 ≤ q ≤ 10^5) — the number of queries.

Each of the next q lines have one of the following formats:

  * 1 t s (1 ≤ t ≤ 10^9, -10^9 ≤ s ≤ 10^9), means that a new event is added, which means that starting from the t-th second the tap's speed will be equal to s. 
  * 2 t (1 ≤ t ≤ 10^9), means that the event which happens at the t-th second must be deleted. Guaranteed that such exists. 
  * 3 l r v (1 ≤ l ≤ r ≤ 10^9, 0 ≤ v ≤ 10^9), means that you should simulate the process from the very beginning of the l-th second till the very beginning of the r-th second inclusive, and to say when will the bowl burst. 



It is guaranteed that t, s, l, r, v in all the queries are integers.

Also, it is guaranteed that there is at least one query of the 3-rd type, and there won't be a query of the 1-st type with such t, that there already exists an event which happens at that second t.

Output

For each query of the 3-rd type, print in a new line the moment when the bowl will burst or print -1 if it won't happen.

Your answer will be considered correct if it's absolute or relative error does not exceed 10^{-6}.

Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} ≤ 10^{-6}.

Examples

Input


6
1 2 1
1 4 -3
3 1 6 1
3 1 6 3
3 1 6 4
3 1 6 5


Output


5
5.666667
6
-1


Input


10
1 2 2
1 4 4
1 7 -10
3 2 4 1
3 5 6 0
3 1 15 1
2 4
3 1 15 1
1 8 1
3 1 15 1


Output


-1
5
8.7
8.1
-1


Input


5
1 1000 9999999
1 2000 -9999
3 1000 2000 0
2 1000
3 1000 2002 1


Output


1000
2000.0001

Note

In the first example all the queries of the 3-rd type cover all the events, it's simulation is following:

<image>
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T &x) {
  x = 0;
  bool sign = false;
  char ch = getchar();
  while (ch < '0' || '9' < ch) {
    sign |= ch == '-';
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  x = sign ? -x : x;
}
template <typename T>
inline void print(T x) {
  static char outp[128];
  int tot = 0;
  if (!x) {
    putchar('0');
    return;
  } else if (x < 0) {
    x = -x;
    putchar('-');
  }
  while (x) {
    outp[tot++] = x % 10, x /= 10;
  }
  while (tot) {
    putchar(outp[--tot] + 48);
  }
}
const long long MAXN = 1e5 + 10;
const long long MAXV = 1e9 + 10;
const long long ERASE = 1008610086;
struct SegmentTree {
  int Size;
  map<long long, long long> Map;
  struct Node {
    long long sum, fcnt;
    int cnt, lson, rson;
    Node() { sum = fcnt = lson = rson = cnt = 0; }
  } tree[MAXN << 6];
  inline void update(int root, int lson, int rson) {
    tree[root].sum = tree[lson].sum + tree[rson].sum;
    tree[root].cnt = tree[lson].cnt + tree[rson].cnt;
    tree[root].fcnt = min(tree[lson].fcnt, tree[lson].sum + tree[rson].fcnt);
  }
  inline void change(int root, int l, int r, int place, long long v) {
    if (l < r) {
      int mid = (l + r) >> 1;
      if (mid >= place) {
        if (!tree[root].lson) {
          tree[root].lson = ++Size;
        }
        change(tree[root].lson, l, mid, place, v);
      } else {
        if (!tree[root].rson) {
          tree[root].rson = ++Size;
        }
        change(tree[root].rson, mid + 1, r, place, v);
      }
      update(root, tree[root].lson, tree[root].rson);
    } else {
      if (v == ERASE) {
        tree[root].sum = tree[root].fcnt = tree[root].cnt = 0;
      } else {
        tree[root].sum = tree[root].fcnt = v;
        tree[root].cnt = 1;
      }
    }
  }
  inline int getcnt(int root, int stdl, int stdr, int l, int r) {
    if (!root || stdl > r || stdr < l) {
      return 0;
    } else if (stdl >= l && stdr <= r) {
      return tree[root].cnt;
    }
    int mid = (stdl + stdr) >> 1;
    return getcnt(tree[root].lson, stdl, mid, l, r) +
           getcnt(tree[root].rson, mid + 1, stdr, l, r);
  }
  long long tot;
  bool flag;
  long double ans;
  inline void query(int root, int stdl, int stdr, int l, int r) {
    if (flag || !root || stdl > r || stdr < l) return;
    if (stdl >= l && stdr <= r) {
      if (tree[root].fcnt + tot <= 0) {
        if (stdl == stdr) {
          int cur = Map[stdl];
          long double dist = (long double)tot / (long double)cur;
          flag = true;
          ans = (long double)stdl - dist;
          tot = 0;
          return;
        } else {
          int mid = (stdl + stdr) >> 1;
          if (tree[tree[root].lson].fcnt + tot <= 0) {
            query(tree[root].lson, stdl, mid, l, r);
          } else {
            tot += tree[tree[root].lson].sum;
            query(tree[root].rson, mid + 1, stdr, l, r);
          }
          return;
        }
      } else {
        tot += tree[root].sum;
        return;
      }
    }
    int mid = (stdl + stdr) >> 1;
    query(tree[root].lson, stdl, mid, l, r);
    query(tree[root].rson, mid + 1, stdr, l, r);
  }
  inline int findrate(long long k) {
    int root = 1;
    int l = 1, r = MAXV;
    while (l < r) {
      int mid = (l + r) / 2;
      if (tree[tree[root].lson].cnt >= k) {
        r = mid;
        root = tree[root].lson;
      } else {
        l = mid + 1;
        k -= tree[tree[root].lson].cnt;
        root = tree[root].rson;
      }
    }
    return l;
  }
  inline void add(long long t, long long v) {
    long long cur = getcnt(1, 1, MAXV, 0, t);
    long long nex = (cur == tree[1].cnt ? MAXV + 1 : findrate(cur + 1));
    cur -= bool(Map.count(t));
    long long frm = (cur == 0 ? -1 : findrate(cur));
    if (frm >= 0) {
      change(1, 1, MAXV, frm, (t - frm) * Map[frm]);
    }
    change(1, 1, MAXV, t, (nex - t) * v);
    Map[t] = v;
  }
  inline void erase(long long t) {
    long long cur = getcnt(1, 1, MAXV, 0, t) - 1;
    long long nex = (cur == tree[1].cnt - 1 ? MAXV + 1 : findrate(cur + 2));
    long long frm = (cur == 0 ? -1 : findrate(cur));
    if (frm >= 0) {
      change(1, 1, MAXV, frm, (nex - frm) * Map[frm]);
    }
    Map.erase(t);
    change(1, 1, MAXV, t, ERASE);
  }
  inline void getans(long long v, int l, int r) {
    tot = v, flag = false, ans = -1;
    query(1, 1, MAXV, l, r);
    if (!flag || r < ans - 1e-8)
      puts("-1");
    else
      cout << fixed << setprecision(9) << ans << '\n';
  }
} Tree;
int main() {
  int n, opt, l, r, time, v;
  read(n);
  Tree.Size = 1;
  while (n--) {
    read(opt);
    if (opt == 1) {
      read(time), read(v);
      Tree.add(time, v);
    } else if (opt == 2) {
      read(time);
      Tree.erase(time);
    } else {
      read(l), read(r), read(v);
      if (v) {
        Tree.getans(v, l, r);
      } else {
        print(l), putchar('\n');
      }
    }
  }
  return 0;
}
