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 + 2;
pair<double, double> merge(pair<double, double> first,
                           pair<double, double> second) {
  if (first.first == -1) return second;
  if (second.first == -1) return first;
  pair<double, double> ret;
  ret.first =
      (first.first * second.first) / (1 - (first.second * (1 - second.first)));
  ret.second =
      second.second + ((1 - second.second) * first.second * second.first) /
                          (1 - (1 - second.first) * first.second);
  return ret;
}
int n;
pair<double, double> t[N * 2];
void build() {
  int i;
  for (i = n - 1; i >= 1; --i) t[i] = merge(t[i << 1], t[i << 1 | 1]);
}
void update(int first, double p) {
  for (first += n - 1, t[first].first = p, t[first].second = p, first >>= 1;
       first >= 1; first >>= 1)
    t[first] = merge(t[first << 1], t[first << 1 | 1]);
}
double query(int l, int r) {
  pair<double, double> first, second;
  first.first = second.first = -1;
  for (l += n - 1, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) first = merge(first, t[l++]);
    if (r & 1) second = merge(t[--r], second);
  }
  return merge(first, second).first;
}
int main() {
  int q, l, ty, r, i;
  cin >> n >> q;
  for (i = 0; i < n; ++i)
    scanf("%d%d", &l, &r), t[i + n].first = t[i + n].second = (double)l / r;
  build();
  while (q--) {
    scanf("%d%d%d", &ty, &i, &l);
    if (ty == 1) {
      scanf("%d", &r);
      update(i, (double)l / r);
    } else
      printf("%0.9lf\n", query(i, l));
  }
}