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 limite = 1000000;
const int tope = 1 << 18;
const int primero = 1 << 17;
void calcula(double d1, double e1, double d2, double e2, double &d, double &e) {
  double p = (e1 * d2) / (1 - e1 * (1 - d2));
  d = d1 * (d2 + (1 - d2) * p);
  e = e2 + (1 - e2) * p;
}
double d[tope];
double e[tope];
void inserta(int pos, double p) {
  pos += primero;
  d[pos] = e[pos] = p;
  while (pos > 1) {
    pos /= 2;
    calcula(d[2 * pos], e[2 * pos], d[2 * pos + 1], e[2 * pos + 1], d[pos],
            e[pos]);
  }
}
double calculaentre(int pos0, int pos1) {
  pos0 += primero;
  pos1 += primero;
  if (pos0 == pos1) return d[pos0];
  double d0 = d[pos0];
  double e0 = e[pos0];
  double d1 = d[pos1];
  double e1 = d[pos1];
  while (pos0 + 1 < pos1) {
    if (pos0 % 2 == 0) calcula(d0, e0, d[pos0 + 1], e[pos0 + 1], d0, e0);
    pos0 /= 2;
    if (pos1 % 2 == 1) calcula(d[pos1 - 1], e[pos1 - 1], d1, e1, d1, e1);
    pos1 /= 2;
  }
  calcula(d0, e0, d1, e1, d0, e0);
  return d0;
}
int n, casos;
int main() {
  ios::sync_with_stdio(false);
  cout.setf(ios::fixed);
  cout.precision(10);
  cin >> n >> casos;
  for (int i = 1; i <= n; i++) {
    int a, b;
    cin >> a >> b;
    inserta(i, double(a) / b);
  }
  for (int cas = 0; cas < casos; cas++) {
    int tipo;
    cin >> tipo;
    if (tipo == 1) {
      int ind, a, b;
      cin >> ind >> a >> b;
      inserta(ind, double(a) / b);
    } else {
      int l, r;
      cin >> l >> r;
      cout << calculaentre(l, r) << endl;
    }
  }
}