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;
constexpr int SIZE = 131072;
double stree1[SIZE * 2 + 1], stree2[SIZE * 2 + 1];
void update_internal(int pos) {
  int child1 = pos << 1;
  int child2 = child1 + 1;
  double a = stree1[child1];
  double b = stree2[child1];
  double c = stree1[child2];
  double d = stree2[child2];
  stree1[pos] = a * c / (1 - (1 - b) * (1 - c));
  stree2[pos] = b * d / (1 - (1 - b) * (1 - c));
}
void update(int pos, int a, int b) {
  pos += SIZE;
  stree1[pos] = (double)a / b;
  stree2[pos] = (double)(b - a) / b;
  for (pos >>= 1; pos >= 1; pos >>= 1) {
    update_internal(pos);
  }
}
double query(int l, int r) {
  vector<pair<double, double>> vt1, vt2;
  l += SIZE;
  r += SIZE;
  while (l <= r) {
    if (l & 1) {
      vt1.emplace_back(stree1[l], stree2[l]);
      l++;
    }
    if ((r & 1) == 0) {
      vt2.emplace_back(stree1[r], stree2[r]);
      r--;
    }
    l >>= 1;
    r >>= 1;
  }
  vt1.insert(vt1.end(), vt2.rbegin(), vt2.rend());
  while (vt1.size() >= 2) {
    auto right = vt1.back();
    vt1.pop_back();
    auto left = vt1.back();
    vt1.pop_back();
    double a = left.first;
    double b = left.second;
    double c = right.first;
    double d = right.second;
    vt1.emplace_back(a * c / (1 - (1 - b) * (1 - c)),
                     b * d / (1 - (1 - b) * (1 - c)));
  }
  return vt1[0].first;
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    stree1[i + SIZE] = (double)a / b;
    stree2[i + SIZE] = (double)(b - a) / b;
  }
  for (int i = n; i < SIZE; i++) {
    stree1[i] = 1;
    stree2[i] = 0;
  }
  for (int i = SIZE - 1; i >= 1; i--) update_internal(i);
  for (int i = 0; i < q; i++) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int idx, a, b;
      scanf("%d%d%d", &idx, &a, &b);
      idx--;
      update(idx, a, b);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      l--;
      r--;
      printf("%.9f\n", query(l, r));
    }
  }
}