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;
template <class X, class Y>
void minimize(X &x, const Y &y) {
  if (x > y) x = y;
}
template <class X, class Y>
void maximize(X &x, const Y &y) {
  if (x < y) x = y;
}
template <class T>
T Abs(const T &x) {
  return (x < 0 ? -x : x);
}
const double eps = 7e-5;
const pair<double, double> NA(-1.0, -1.0);
class SegmentTree {
 private:
  int n;
  vector<pair<double, double> > tree;
  static pair<double, double> concat(const pair<double, double> &___left,
                                     const pair<double, double> &___right) {
    if (___left.first < 0) return (___right);
    if (___right.first < 0) return (___left);
    double A = ___right.first /
               (1.0 - ___left.second + ___right.first * ___left.second);
    double B = ___right.first * ___left.second /
               (1.0 - ___left.second + ___right.first * ___left.second);
    double f = ___left.first * A;
    double g = ___right.second + (1.0 - ___right.second) * B;
    return (make_pair(f, g));
  }
  void update(int i, int l, int r, int x, double v) {
    if (l == r) {
      tree[i] = make_pair(v, v);
      return;
    }
    int m = (l + r) >> 1;
    if (x > m)
      update(2 * i + 1, m + 1, r, x, v);
    else
      update(2 * i, l, m, x, v);
    tree[i] = concat(tree[2 * i], tree[2 * i + 1]);
  }
  pair<double, double> get(int i, int l, int r, int u, int v) const {
    if (l > v || r < u || l > r || v < u) return (NA);
    if (u <= l && r <= v) return (tree[i]);
    int m = (l + r) >> 1;
    pair<double, double> L = get(2 * i, l, m, u, v);
    pair<double, double> R = get(2 * i + 1, m + 1, r, u, v);
    return (concat(L, R));
  }

 public:
  SegmentTree(int n = 0) {
    this->n = n;
    if (n > 0) tree.assign(4 * n + 7, NA);
  }
  void update(int x, double v) { update(1, 1, n, x, v); }
  double get(int l, int r) const { return (get(1, 1, n, l, r).first); }
};
void process(void) {
  int n, q;
  scanf("%d%d", &n, &q);
  SegmentTree myit(n);
  for (int i = (1), _b = (n); i <= _b; i = i + 1) {
    int a, b;
    scanf("%d%d", &a, &b);
    myit.update(i, 1.0 * a / b);
  }
  cerr << "DONE" << endl;
  for (int love = 0, _n = (q); love < _n; love = love + 1) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int x, a, b;
      scanf("%d%d%d", &x, &a, &b);
      myit.update(x, 1.0 * a / b);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      printf("%.7lf\n", myit.get(l, r));
    }
  }
}
int main(void) {
  process();
  return 0;
}