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>
inline int read() {
  int s = 0;
  char c;
  while ((c = getchar()) < '0' || c > '9')
    ;
  do {
    s = s * 10 + c - '0';
  } while ((c = getchar()) >= '0' && c <= '9');
  return s;
}
const int N = 100010;
double tl[262144], tr[262144];
int S, n, m;
double query(int l, int r) {
  double l1 = 1, l2 = 1, r1 = 0, r2 = 0, lx, rx;
  for (l = l + S - 1, r = r + S + 1; l ^ r ^ 1; l >>= 1, r >>= 1) {
    if (~l & 1)
      lx = ((l1) * (tl[l ^ 1]) / ((1 - (r1) * (1 - (tl[l ^ 1]))))),
      rx = ((tr[l ^ 1]) + (1 - (tr[l ^ 1])) * (r1) * (tl[l ^ 1]) /
                              (1 - (r1) * (1 - (tl[l ^ 1])))),
      l1 = lx, r1 = rx;
    if (r & 1)
      lx = ((tl[r ^ 1]) * (l2) / ((1 - (tr[r ^ 1]) * (1 - (l2))))),
      rx = ((r2) +
            (1 - (r2)) * (tr[r ^ 1]) * (l2) / (1 - (tr[r ^ 1]) * (1 - (l2)))),
      l2 = lx, r2 = rx;
  }
  return ((l1) * (l2) / ((1 - (r1) * (1 - (l2)))));
}
int main() {
  int i;
  n = read();
  m = read();
  for (S = 1; S <= n + 2; S <<= 1)
    ;
  for (i = n + 1; i < S; i++) tl[S + i] = tr[S + i] = 1;
  for (i = 1; i <= n; i++) {
    double a = read();
    tl[S + i] = tr[S + i] = a / (double)read();
  }
  for (i = S - 1; i; i--) {
    tl[i] = ((tl[i + i]) * (tl[i + i + 1]) /
             ((1 - (tr[i + i]) * (1 - (tl[i + i + 1]))))),
    tr[i] = ((tr[i + i + 1]) + (1 - (tr[i + i + 1])) * (tr[i + i]) *
                                   (tl[i + i + 1]) /
                                   (1 - (tr[i + i]) * (1 - (tl[i + i + 1]))));
  };
  while (m--) {
    if (read() == 1) {
      int tg = read() + S;
      double p = read();
      p /= (double)read();
      tl[tg] = tr[tg] = p;
      for (tg >>= 1; tg; tg >>= 1) {
        tl[tg] = ((tl[tg + tg]) * (tl[tg + tg + 1]) /
                  ((1 - (tr[tg + tg]) * (1 - (tl[tg + tg + 1]))))),
        tr[tg] = ((tr[tg + tg + 1]) +
                  (1 - (tr[tg + tg + 1])) * (tr[tg + tg]) * (tl[tg + tg + 1]) /
                      (1 - (tr[tg + tg]) * (1 - (tl[tg + tg + 1]))));
      };
    } else {
      int l = read(), r = read();
      printf("%.10lf\n", query(l, r));
    }
  }
  return 0;
}