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>
#pragma comment(linker, "/STACK:200000000")
using namespace std;
struct _ {
  _() {
    cout.precision(30);
    ios_base::sync_with_stdio(0);
  }
} __;
const int N = 1e5 + 10;
const double EPS = 1e-6;
struct segment {
  double d, kr;
};
void check(segment s) {
  if (!(0 <= s.d && s.d <= 1.0)) {
    cout << "Error";
  }
  if (!(0 <= s.kr && s.kr <= 1.0)) {
    cout << "Error";
  }
}
segment operator+(segment left, segment right) {
  segment result;
  result.kr = right.kr + (1 - right.kr) * left.kr * right.d /
                             (1 - (1 - right.d) * left.kr);
  result.d = left.d * right.d / (1 - (1 - right.d) * left.kr);
  return result;
}
struct sTree {
  segment t[4 * N];
  void set(int v, int tl, int tr, int ix, double value) {
    if (tl == tr) {
      t[v].d = t[v].kr = value;
      return;
    }
    int tm = (tl + tr) / 2;
    if (ix <= tm) {
      set(2 * v + 1, tl, tm, ix, value);
    } else {
      set(2 * v + 2, tm + 1, tr, ix, value);
    }
    t[v] = t[2 * v + 1] + t[2 * v + 2];
  }
  segment get(int v, int tl, int tr, int l, int r) {
    if (l == tl && r == tr) {
      return t[v];
    }
    int tm = (tl + tr) / 2;
    segment a, b;
    bool x = false, y = false;
    if (l <= min(tm, r)) {
      a = get(2 * v + 1, tl, tm, l, min(tm, r));
      x = true;
    }
    if (max(tm + 1, l) <= r) {
      b = get(2 * v + 2, tm + 1, tr, max(tm + 1, l), r);
      y = true;
    }
    if (!x) return b;
    if (!y) return a;
    return a + b;
  }
} tree;
int main() {
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    int a, b;
    cin >> a >> b;
    tree.set(0, 0, n - 1, i, a * 1.0l / b);
  }
  for (int i = 0; i < q; i++) {
    int type;
    cin >> type;
    if (type == 1) {
      int ix, a, b;
      cin >> ix >> a >> b;
      ix--;
      tree.set(0, 0, n - 1, ix, a * 1.0l / b);
    } else {
      int l, r;
      cin >> l >> r;
      l--, r--;
      segment s = tree.get(0, 0, n - 1, l, r);
      cout << s.d << endl;
    }
  }
  return 0;
}