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 maxn = 100005;
double a[maxn];
struct pi {
  int le, ri;
  double l, r;
} pp[maxn << 2];
void update(int tot) {
  double l1 = pp[2 * tot].l;
  double l2 = pp[2 * tot + 1].l;
  double r1 = pp[2 * tot].r;
  double r2 = pp[2 * tot + 1].r;
  pp[tot].l = l1 * l2 / (1 + r1 * (l2 - 1));
  pp[tot].r = r2 + (r1 * l2 * (1 - r2) / (1 - r1 * (1 - l2)));
}
void build(int tot, int l, int r) {
  pp[tot].le = l;
  pp[tot].ri = r;
  if (l == r) {
    pp[tot].l = a[l];
    pp[tot].r = a[l];
    return;
  }
  build(2 * tot, l, (l + r) / 2);
  build(2 * tot + 1, (l + r) / 2 + 1, r);
  update(tot);
}
void merg(int tot, int x, double p) {
  if (pp[tot].le == pp[tot].ri) {
    pp[tot].l = pp[tot].r = p;
    return;
  }
  int mid = (pp[tot].le + pp[tot].ri) / 2;
  if (x <= mid)
    merg(2 * tot, x, p);
  else
    merg(2 * tot + 1, x, p);
  update(tot);
}
pair<double, double> query(int tot, int l, int r) {
  if (pp[tot].le >= l && pp[tot].ri <= r) {
    return make_pair(pp[tot].l, pp[tot].r);
  }
  int mid = (pp[tot].le + pp[tot].ri) / 2;
  if (r <= mid)
    return query(2 * tot, l, r);
  else if (l > mid)
    return query(2 * tot + 1, l, r);
  else {
    pair<double, double> x = query(2 * tot, l, r);
    pair<double, double> y = query(2 * tot + 1, l, r);
    double l1 = x.first;
    double l2 = y.first;
    double r1 = x.second;
    double r2 = y.second;
    double a = l1 * l2 / (1 + r1 * (l2 - 1));
    double b = r2 + (r1 * l2 * (1 - r2) / (1 - r1 * (1 - l2)));
    return make_pair(a, b);
  }
}
int main() {
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    int x, b;
    scanf("%d%d", &x, &b);
    a[i] = 1.0 * x / b;
  }
  build(1, 1, n);
  for (int i = 0; i < q; i++) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int a, b, p;
      scanf("%d%d%d", &p, &a, &b);
      merg(1, p, 1.0 * a / b);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      printf("%.10lf\n", query(1, l, r).first);
    }
  }
}