Problem: There are n casinos lined in a row. If Memory plays at casino i, he has probability pi to win and move to the casino on the right (i + 1) or exit the row (if i = n), and a probability 1 - pi to lose and move to the casino on the left (i - 1) or also exit the row (if i = 1). 

We say that Memory dominates on the interval i... j if he completes a walk such that,

  * He starts on casino i. 
  * He never looses in casino i. 
  * He finishes his walk by winning in casino j. 



Note that Memory can still walk left of the 1-st casino and right of the casino n and that always finishes the process.

Now Memory has some requests, in one of the following forms:

  * 1 i a b: Set <image>. 
  * 2 l r: Print the probability that Memory will dominate on the interval l... r, i.e. compute the probability that Memory will first leave the segment l... r after winning at casino r, if she starts in casino l. 



It is guaranteed that at any moment of time p is a non-decreasing sequence, i.e. pi ≤ pi + 1 for all i from 1 to n - 1.

Please help Memory by answering all his requests!

Input

The first line of the input contains two integers n and q(1 ≤ n, q ≤ 100 000), — number of casinos and number of requests respectively.

The next n lines each contain integers ai and bi (1 ≤ ai < bi ≤ 109) — <image> is the probability pi of winning in casino i.

The next q lines each contain queries of one of the types specified above (1 ≤ a < b ≤ 109, 1 ≤ i ≤ n, 1 ≤ l ≤ r ≤ n).

It's guaranteed that there will be at least one query of type 2, i.e. the output will be non-empty. Additionally, it is guaranteed that p forms a non-decreasing sequence at all times.

Output

Print a real number for every request of type 2 — the probability that boy will "dominate" on that interval. Your answer will be considered correct if its absolute error does not exceed 10 - 4.

Namely: let's assume that one of your answers is a, and the corresponding answer of the jury is b. The checker program will consider your answer correct if |a - b| ≤ 10 - 4.

Example

Input

3 13
1 3
1 2
2 3
2 1 1
2 1 2
2 1 3
2 2 2
2 2 3
2 3 3
1 2 2 3
2 1 1
2 1 2
2 1 3
2 2 2
2 2 3
2 3 3


Output

0.3333333333
0.2000000000
0.1666666667
0.5000000000
0.4000000000
0.6666666667
0.3333333333
0.2500000000
0.2222222222
0.6666666667
0.5714285714
0.6666666667

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct Segment {
  double LR, LL, RL, RR;
  static Segment casino(const double p) { return {p, 1 - p, 1 - p, p}; }
  static Segment unit() { return {1, 0, 1, 0}; }
};
Segment operator*(const Segment a, const Segment b) {
  const auto LR = (a.LR * b.LR) / (1 - b.LL * a.RR);
  const auto LL = 1 - LR;
  const auto RL = (b.RL * a.RL) / (1 - a.RR * b.LL);
  const auto RR = 1 - RL;
  return {LR, LL, RL, RR};
}
struct SegmentTree {
  int N;
  vector<Segment> data;
  SegmentTree(int N, const vector<double>& initial) : N(N), data(2 * N) {
    for (auto a = 0; a < N; a++) data[N + a] = Segment::casino(initial[a]);
    for (auto i = N - 1; i >= 1; i--) data[i] = data[2 * i] * data[2 * i + 1];
  }
  void update(const int a, const double p) {
    auto i = N + a;
    data[i] = Segment::casino(p);
    for (i /= 2; i >= 1; i /= 2) data[i] = data[2 * i] * data[2 * i + 1];
  }
  double query(const int a, const int b) const {
    auto left = Segment::unit(), right = Segment::unit();
    for (auto i = a + N, j = b + N + 1; i < j; i /= 2, j /= 2) {
      if (i % 2) left = left * data[i++];
      if (j % 2) right = data[--j] * right;
    }
    return (left * right).LR;
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, q;
  cin >> n >> q;
  auto initial = vector<double>();
  for (auto i = 0; i < n; i++) {
    int a, b;
    cin >> a >> b;
    initial.push_back(static_cast<double>(a) / b);
  }
  auto tree = SegmentTree(n, initial);
  for (auto i = 0; i < q; i++) {
    int type;
    cin >> type;
    if (type == 1) {
      int i, a, b;
      cin >> i >> a >> b;
      i--;
      const auto p = static_cast<double>(a) / b;
      tree.update(i, p);
    } else if (type == 2) {
      int l, r;
      cin >> l >> r;
      l--;
      r--;
      cout << tree.query(l, r) << "\n";
    } else
      throw invalid_argument("Invalid operation type: " + to_string(type));
  }
}