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;
double pr[100000];
pair<double, double> seg[400000];
bool ze(double d) { return (abs(d) <= 0.0000000001); }
pair<double, double> merge(pair<double, double> a, pair<double, double> b) {
  if (ze(a.first + 1) && ze(a.second + 1)) return b;
  if (ze(b.first + 1) && ze(b.second + 1)) return a;
  double l1 = a.first;
  double r1 = a.second;
  double l2 = b.first;
  double r2 = b.second;
  if (ze(((l2 - 1) * r1 + 1))) return make_pair(0, 0);
  double le = l1 * l2 / ((l2 - 1) * r1 + 1);
  double ri = r2 + (r1 * l2 * (-r2 + 1)) / (-r1 * (-l2 + 1) + 1);
  return make_pair(le, ri);
}
void build(int no, int b, int e) {
  if (b == e) {
    seg[no] = make_pair(pr[b], pr[b]);
    return;
  }
  int mid = (b + e) / 2;
  build(2 * no, b, mid);
  build(2 * no + 1, mid + 1, e);
  seg[no] = merge(seg[2 * no], seg[2 * no + 1]);
}
void upd(int no, int b, int e, int i, double val) {
  if (i < b || i > e) return;
  if (b == e) {
    seg[no] = make_pair(val, val);
    return;
  }
  int mid = (b + e) / 2;
  upd(2 * no, b, mid, i, val);
  upd(2 * no + 1, mid + 1, e, i, val);
  seg[no] = merge(seg[2 * no], seg[2 * no + 1]);
}
pair<double, double> query(int no, int b, int e, int l, int r) {
  if (b > r || e < l) return make_pair(-1, -1);
  if (l <= b && e <= r) return seg[no];
  int mid = (b + e) / 2;
  return merge(query(2 * no, b, mid, l, r),
               query(2 * no + 1, mid + 1, e, l, r));
}
int main() {
  int q;
  scanf("%d %d", &n, &q);
  for (int i = 0; i < n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    pr[i] = (double)a / (double)b;
  }
  build(1, 0, n - 1);
  for (int que = 0; que < q; que++) {
    int ty;
    int a;
    int b;
    scanf("%d %d %d", &ty, &a, &b);
    if (ty == 1) {
      int c;
      scanf("%d", &c);
      upd(1, 0, n - 1, a - 1, (double)b / (double)c);
    } else {
      printf("%.20f\n", query(1, 0, n - 1, a - 1, b - 1).first);
    }
  }
  return 0;
}