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 = 1e5 + 100;
const int B = 333;
struct node {
  double pl, pr;
  double rpl, rpr;
};
node sum[N << 2];
int a[N], b[N];
int n, q;
node solve(node a, node b) {
  node now;
  now.pr = a.pr / (1 - a.rpr * b.pl) * b.pr;
  now.pl = 1 - now.pr;
  now.rpl = b.rpl / (1 - a.rpr * b.pl) * a.rpl;
  now.rpr = 1 - now.rpl;
  return now;
}
void pushup(int rt) { sum[rt] = solve(sum[rt << 1], sum[rt << 1 | 1]); }
void build(int l, int r, int rt) {
  if (l == r) {
    double now = a[l] * 1.0 / b[l];
    sum[rt].pl = 1 - now;
    sum[rt].pr = now;
    sum[rt].rpl = 1 - now;
    sum[rt].rpr = now;
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, rt << 1);
  build(mid + 1, r, rt << 1 | 1);
  pushup(rt);
}
node query(int L, int R, int l, int r, int rt) {
  if (L <= l && R >= r) {
    return sum[rt];
  }
  int mid = (l + r) >> 1;
  node now{0, 1};
  if (mid < R) now = query(L, R, mid + 1, r, rt << 1 | 1);
  if (mid >= L) now = solve(query(L, R, l, mid, rt << 1), now);
  return now;
}
void update(int x, double f, int l, int r, int rt) {
  if (l == r) {
    sum[rt].pr = f;
    sum[rt].pl = 1 - f;
    sum[rt].rpr = f;
    sum[rt].rpl = 1 - f;
    return;
  }
  int mid = (l + r) >> 1;
  if (mid >= x)
    update(x, f, l, mid, rt << 1);
  else
    update(x, f, mid + 1, r, rt << 1 | 1);
  pushup(rt);
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &a[i], &b[i]);
  }
  build(1, n, 1);
  node a = sum[4], b = sum[5], c = sum[3];
  node now = solve(b, c);
  now = solve(a, now);
  for (int i = 1; i <= q; i++) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int x, a, b;
      scanf("%d %d %d", &x, &a, &b);
      update(x, a * 1.0 / b, 1, n, 1);
    } else {
      int l, r;
      scanf("%d %d", &l, &r);
      node ans = query(l, r, 1, n, 1);
      printf("%.10f\n", ans.pr);
    }
  }
  return 0;
}