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 + 7;
int n, q;
pair<double, double> p[N << 2];
pair<double, double> up(pair<double, double> a, pair<double, double> b) {
  double f, s;
  double q = (1 - b.first) * a.second;
  double a0 = a.first * b.first;
  f = a0 / (1 - q);
  q = (1 - b.first) * a.second;
  a0 = (1 - b.second) * a.second * b.first;
  s = a0 / (1 - q);
  return make_pair(f, s + b.second);
}
void build(int l, int r, int rt) {
  if (l == r) {
    double u, d;
    scanf("%lf%lf", &u, &d);
    p[rt].first = p[rt].second = u / d;
    return;
  }
  int mid = l + r >> 1;
  build(l, mid, rt << 1);
  build(mid + 1, r, rt << 1 | 1);
  p[rt] = up(p[rt << 1], p[rt << 1 | 1]);
}
void update(int l, int r, int rt, int pos, double u, double d) {
  if (l == r) {
    p[rt].first = p[rt].second = u / d;
    return;
  }
  int mid = l + r >> 1;
  if (pos <= mid)
    update(l, mid, rt << 1, pos, u, d);
  else
    update(mid + 1, r, rt << 1 | 1, pos, u, d);
  p[rt] = up(p[rt << 1], p[rt << 1 | 1]);
}
pair<double, double> query(int l, int r, int rt, int L, int R) {
  if (L <= l && r <= R) {
    return p[rt];
  }
  int mid = l + r >> 1;
  if (R <= mid)
    return query(l, mid, rt << 1, L, R);
  else if (L > mid)
    return query(mid + 1, r, rt << 1 | 1, L, R);
  else
    return up(query(l, mid, rt << 1, L, R),
              query(mid + 1, r, rt << 1 | 1, L, R));
}
int main() {
  cin >> n >> q;
  build(1, n, 1);
  for (int i = 1; i <= q; ++i) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int pos;
      double u, d;
      scanf("%d%lf%lf", &pos, &u, &d);
      update(1, n, 1, pos, u, d);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      auto it = query(1, n, 1, l, r);
      printf("%.10f\n", it.first);
    }
  }
  return 0;
}