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 Maxn = 100005;
int n, q;
struct Tree {
  int lt, rt;
  double prod, sum;
} tree[4 * Maxn];
void build(int root, int lt, int rt) {
  tree[root] = (Tree){lt, rt, 0, 0};
  if (lt + 1 != rt) {
    int mid = (lt + rt) >> 1;
    build(root << 1, lt, mid);
    build(root << 1 | 1, mid, rt);
  }
}
void modify(int root, int pos, double val) {
  if (tree[root].lt + 1 == tree[root].rt)
    tree[root].prod = tree[root].sum = val;
  else {
    int mid = (tree[root].lt + tree[root].rt) >> 1;
    if (pos >= mid)
      modify(root << 1 | 1, pos, val);
    else
      modify(root << 1, pos, val);
    tree[root].prod = tree[root << 1].prod * tree[root << 1 | 1].prod;
    tree[root].sum =
        tree[root << 1].sum + tree[root << 1].prod * tree[root << 1 | 1].sum;
  }
}
pair<double, double> ask(int root, int lt, int rt) {
  if (tree[root].lt == lt && tree[root].rt == rt)
    return make_pair(tree[root].prod, tree[root].sum);
  else {
    int mid = (tree[root].lt + tree[root].rt) >> 1;
    if (lt >= mid)
      return ask(root << 1 | 1, lt, rt);
    else if (rt <= mid)
      return ask(root << 1, lt, rt);
    else {
      pair<double, double> tmp1 = ask(root << 1, lt, mid),
                           tmp2 = ask(root << 1 | 1, mid, rt);
      return make_pair(tmp1.first * tmp2.first,
                       tmp1.second + tmp1.first * tmp2.second);
    }
  }
}
int main() {
  scanf("%d%d", &n, &q);
  build(1, 1, n + 1);
  for (int i = 1; i <= n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    double w = x / (double)y;
    modify(1, i, (1 - w) / w);
  }
  while (q--) {
    int opt;
    scanf("%d", &opt);
    if (opt == 1) {
      int pos, x, y;
      scanf("%d%d%d", &pos, &x, &y);
      double w = x / (double)y;
      modify(1, pos, (1 - w) / w);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      printf("%.10lf\n", 1 / (1 + ask(1, l, r + 1).second));
    }
  }
  return 0;
}