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;
const int N = 1e5 + 10;
int n, q;
pair<long double, long double> seg[4 * N];
pair<long double, long double> merge(pair<long double, long double> p,
                                     pair<long double, long double> q) {
  if (p.first == -1) return q;
  if (q.first == -1) return p;
  long double Fi = (p.first * q.first) / (1.0 - p.second * (1.0 - q.first));
  long double Se = q.second + (p.second * q.first * (1.0 - q.second)) /
                                  (1 - p.second * (1.0 - q.first));
  return {Fi, Se};
}
void upd(int p, long double x, int id = 1, int s = 0, int e = n) {
  if (e - s < 2) {
    seg[id] = {x, x};
    return;
  }
  int mid = (s + e) / 2;
  if (p < mid)
    upd(p, x, 2 * id, s, mid);
  else
    upd(p, x, 2 * id + 1, mid, e);
  seg[id] = merge(seg[2 * id], seg[2 * id + 1]);
}
pair<long double, long double> get(int L, int R, int id = 1, int s = 0,
                                   int e = n) {
  if (L <= s && e <= R) return seg[id];
  if (L >= e || s >= R) return {-1, -1};
  int mid = (s + e) / 2;
  return merge(get(L, R, 2 * id, s, mid), get(L, R, 2 * id + 1, mid, e));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    int s, m;
    cin >> s >> m;
    upd(i, (long double)(s) / (long double)(m));
  }
  cout << fixed << setprecision(10);
  for (int i = 0; i < q; i++) {
    int tp;
    cin >> tp;
    if (tp == 1) {
      int p, s, m;
      cin >> p >> s >> m;
      --p;
      upd(p, (long double)(s) / (long double)(m));
    } else {
      int l, r;
      cin >> l >> r;
      --l;
      cout << get(l, r).first << '\n';
    }
  }
}