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;
int n, m;
struct node {
  pair<double, double> f;
  int lb, rb;
} tr[2000000];
pair<double, double> merge(pair<double, double> l, pair<double, double> r) {
  double a2 = r.first / (1 - l.second + l.second * r.first), a1 = l.second * a2;
  return make_pair(l.first * a2, r.second + (1 - r.second) * a1);
}
void change(int t, int p, double d) {
  if (p == tr[t].lb && p == tr[t].rb) {
    tr[t].f = make_pair(d, d);
    return;
  }
  if (p <= tr[t * 2].rb) change(t * 2, p, d);
  if (p >= tr[t * 2 + 1].lb) change(t * 2 + 1, p, d);
  tr[t].f = merge(tr[t * 2].f, tr[t * 2 + 1].f);
}
pair<double, double> get_ans(int t, int l, int r) {
  if (l == tr[t].lb && r == tr[t].rb) return tr[t].f;
  pair<double, double> s = make_pair(1, 1);
  if (r <= tr[t * 2].rb)
    s = get_ans(t * 2, l, min(tr[t * 2].rb, r));
  else if (l >= tr[t * 2 + 1].lb)
    s = get_ans(t * 2 + 1, max(tr[t * 2 + 1].lb, l), r);
  else
    s = merge(get_ans(t * 2, l, min(tr[t * 2].rb, r)),
              get_ans(t * 2 + 1, max(tr[t * 2 + 1].lb, l), r));
  return s;
}
void build(int t, int l, int r) {
  tr[t].lb = l;
  tr[t].rb = r;
  if (l == r) {
    int p, q;
    cin >> p >> q;
    tr[t].f = make_pair(1.0 * p / q, 1.0 * p / q);
  } else {
    build(t * 2, l, (l + r) / 2);
    build(t * 2 + 1, (l + r) / 2 + 1, r);
    tr[t].f = merge(tr[t * 2].f, tr[t * 2 + 1].f);
  }
}
int main() {
  cin >> n >> m;
  build(1, 1, n);
  for (int i = 1; i <= m; ++i) {
    int opt, a, b, c;
    cin >> opt >> a >> b;
    if (opt == 1) {
      cin >> c;
      change(1, a, 1.0 * b / c);
    } else {
      cout << get_ans(1, a, b).first << "\n";
    }
  }
}