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;
const int N = 100005;
int n, q;
double u[N], tich[4 * N], tong[4 * N];
void input() {
  int a, b;
  scanf("%d%d", &n, &q);
  for (int i = (1); i <= (n); i++) {
    scanf("%d%d", &a, &b);
    u[i] = (double)(b) / a - 1;
  }
}
void build(int i, int l, int r) {
  if (l > r) return;
  if (l == r) {
    tich[i] = u[l];
    tong[i] = u[l];
    return;
  }
  int mid = (l + r) / 2;
  build(i * 2, l, mid);
  build(i * 2 + 1, mid + 1, r);
  tich[i] = tich[i * 2] * tich[i * 2 + 1];
  tong[i] = tong[i * 2] + tich[i * 2] * tong[i * 2 + 1];
}
void up(int k, int l, int r, int x, double u) {
  if (x > r || x < l) return;
  if (l == r) {
    tich[k] = u;
    tong[k] = u;
    return;
  }
  int mid = (l + r) / 2;
  up(k * 2, l, mid, x, u);
  up(k * 2 + 1, mid + 1, r, x, u);
  tich[k] = tich[k * 2] * tich[k * 2 + 1];
  tong[k] = tong[k * 2] + tich[k * 2] * tong[k * 2 + 1];
}
double get1(int k, int l, int r, int x, int y) {
  if (y < l || x > r) return 1;
  if (x <= l && r <= y) return tich[k];
  int mid = (l + r) / 2;
  return get1(k * 2, l, mid, x, y) * get1(k * 2 + 1, mid + 1, r, x, y);
}
double get2(int k, int l, int r, int x, int y) {
  if (l >= x && r <= y) return tong[k];
  int mid = (l + r) / 2;
  if (mid < x) return get2(k * 2 + 1, mid + 1, r, x, y);
  if (mid >= y) return get2(k * 2, l, mid, x, y);
  return get2(k * 2, l, mid, x, mid) +
         get1(1, 1, n, x, mid) * get2(k * 2 + 1, mid + 1, r, mid + 1, y);
}
void solve() {
  int t, y, a, b, x;
  while (q--) {
    scanf("%d", &t);
    if (t == 2) {
      scanf("%d%d", &x, &y);
      double kq = get2(1, 1, n, x, y);
      if (kq < 1E30)
        kq = (1.0) / (1.0 + kq);
      else
        kq = 0;
      cout << kq << endl;
    } else {
      scanf("%d%d%d", &x, &a, &b);
      double u = (double)(b) / a - 1;
      up(1, 1, n, x, u);
    }
  }
}
int main() {
  input();
  build(1, 1, n);
  solve();
}