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;
int n, m;
double u[100010];
struct node {
  double prod, val;
};
node merge(node a, node b) {
  double fr, sc;
  if (a.prod < 1e-12 or b.prod < 1e-12) {
    fr = 0.0;
  } else {
    fr = a.prod * b.prod;
  }
  if (a.prod < 1e-12 or b.val < 1e-12) {
    sc = 0.0;
  } else {
    sc = a.prod * b.val;
  }
  return node{fr, a.val + sc};
}
int getmid(int l, int r) { return l + (r - l) / 2; }
struct segtree {
  node ar[400010];
  void init(int idx, int l, int r) {
    if (l == r) {
      ar[idx].prod = u[l];
      ar[idx].val = u[l];
      return;
    }
    int mid = getmid(l, r);
    init(idx * 2 + 1, l, mid);
    init(idx * 2 + 2, mid + 1, r);
    ar[idx] = merge(ar[idx * 2 + 1], ar[idx * 2 + 2]);
  }
  void upd(int idx, int l, int r, int x, double newval) {
    if (l == r) {
      ar[idx].prod = newval;
      ar[idx].val = newval;
      return;
    }
    int mid = getmid(l, r);
    if (x <= mid) {
      upd(idx * 2 + 1, l, mid, x, newval);
    } else {
      upd(idx * 2 + 2, mid + 1, r, x, newval);
    }
    ar[idx] = merge(ar[idx * 2 + 1], ar[idx * 2 + 2]);
  }
  node query(int idx, int l, int r, int ql, int qr) {
    if (l > qr or r < ql) {
      return node{1.0, 0.0};
    }
    if (ql <= l and qr >= r) {
      return ar[idx];
    }
    int mid = getmid(l, r);
    return merge(query(idx * 2 + 1, l, mid, ql, qr),
                 query(idx * 2 + 2, mid + 1, r, ql, qr));
  }
};
segtree tree;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    int a, b;
    cin >> a >> b;
    double p = (0.0 + a) / b;
    u[i] = (1.0 - p) / p;
  }
  tree.init(0, 0, n - 1);
  for (int i = 0; i < m; ++i) {
    int x, idx, a, b;
    cin >> x;
    if (x == 1) {
      cin >> idx >> a >> b;
      double p = (1.0 * a) / b;
      tree.upd(0, 0, n - 1, idx - 1, (1 - p) / p);
    } else {
      cin >> a >> b;
      node tmp = tree.query(0, 0, n - 1, a - 1, b - 1);
      double ans = (1.0 + tmp.val);
      cout << setprecision(10) << fixed << 1.0 / ans << "\n";
    }
  }
}