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;
double base[100005];
struct Seg {
  vector<pair<double, double> > tree;
  Seg(int V) : tree(4 * V + 2) {}
  void build(int x, int L, int R) {
    if (L == R) {
      tree[x].first = base[L];
      tree[x].second = 1.0 - base[L];
      return;
    }
    int mid = (L + R) >> 1;
    build(x + x, L, mid);
    build(x + x + 1, mid + 1, R);
    pair<double, double> p = tree[x + x], q = tree[x + x + 1];
    tree[x].first =
        p.first * q.first * (1.0 / (1.0 - (1.0 - q.first) * (1.0 - p.second)));
    tree[x].second = p.second * q.second *
                     (1.0 / (1.0 - (1.0 - p.second) * (1.0 - q.first)));
  }
  void update(int x, int L, int R, int idx, double per) {
    if (idx < L || R < idx) return;
    if (L == R && L == idx) {
      tree[x].first = per;
      tree[x].second = 1.0 - per;
      return;
    }
    int mid = (L + R) >> 1;
    update(x + x, L, mid, idx, per);
    update(x + x + 1, mid + 1, R, idx, per);
    pair<double, double> p = tree[x + x];
    pair<double, double> q = tree[x + x + 1];
    double r1 =
        p.first * q.first * (1.0 / (1.0 - (1.0 - q.first) * (1.0 - p.second)));
    double r2 = p.second * q.second *
                (1.0 / (1.0 - (1.0 - p.second) * (1.0 - q.first)));
    tree[x] = make_pair(r1, r2);
  }
  pair<double, double> domi(int x, int L, int R, int left, int right) {
    if (R < left || L > right) {
      return make_pair(1.0, 1.0);
    }
    if (left <= L && R <= right) {
      return tree[x];
    }
    int mid = (L + R) >> 1;
    pair<double, double> p = domi(x + x, L, mid, left, right);
    pair<double, double> q = domi(x + x + 1, mid + 1, R, left, right);
    double r1 =
        p.first * q.first * (1.0 / (1.0 - (1.0 - q.first) * (1.0 - p.second)));
    double r2 = p.second * q.second *
                (1.0 / (1.0 - (1.0 - p.second) * (1.0 - q.first)));
    return make_pair(r1, r2);
  }
};
int n, query;
int main() {
  scanf("%d %d", &n, &query);
  for (int a, b, i = 1; i <= n; i++) {
    scanf("%d %d", &a, &b);
    base[i] = (double)a / (double)b;
  }
  Seg seg(n);
  seg.build(1, 1, n);
  for (int ty, a, b, c, i = 1; i <= query; i++) {
    scanf("%d", &ty);
    if (ty == 1) {
      scanf("%d %d %d", &a, &b, &c);
      seg.update(1, 1, n, a, (double)b / (double)c);
    } else {
      scanf("%d %d", &a, &b);
      double key = seg.domi(1, 1, n, a, b).first;
      printf("%.6lf\n", key);
    }
  }
}