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 = 1e5 + 7;
struct treenode {
  int L, R;
  pair<double, double> A;
  void updata(pair<double, double> v) { A = v; }
};
pair<double, double> tmp = make_pair(1.0, 0);
treenode tree[maxn * 4];
inline void push_up(int o) {
  tree[o].A.first = tree[o * 2].A.first * tree[o * 2 + 1].A.first;
  tree[o].A.second =
      tree[o * 2].A.second + tree[o * 2 + 1].A.second * tree[o * 2].A.first;
}
inline void build_tree(int L, int R, int o) {
  tree[o].L = L, tree[o].R = R;
  if (R > L) {
    int mid = (L + R) >> 1;
    build_tree(L, mid, o * 2);
    build_tree(mid + 1, R, o * 2 + 1);
  }
}
inline void update(int QL, int QR, pair<double, double> v, int o) {
  int L = tree[o].L, R = tree[o].R;
  if (QL <= L && R <= QR)
    tree[o].updata(v);
  else {
    int mid = (L + R) >> 1;
    if (QL <= mid) update(QL, QR, v, o * 2);
    if (QR > mid) update(QL, QR, v, o * 2 + 1);
    push_up(o);
  }
}
inline pair<double, double> query(int QL, int QR, int o) {
  int L = tree[o].L, R = tree[o].R;
  if (QL <= L && R <= QR)
    return tree[o].A;
  else {
    int mid = (L + R) >> 1;
    pair<double, double> AA = tmp, BB = tmp, CC;
    if (QL <= mid) AA = query(QL, QR, 2 * o);
    if (QR > mid) BB = query(QL, QR, 2 * o + 1);
    push_up(o);
    CC.first = AA.first * BB.first;
    CC.second = AA.second + BB.second * AA.first;
    return CC;
  }
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  build_tree(1, n, 1);
  for (int i = 1; i <= n; i++) {
    double a, b;
    cin >> a >> b;
    double p = a / b;
    update(i, i, make_pair((1.0 - p) / p, (1.0 - p) / p), 1);
  }
  for (int i = 1; i <= q; i++) {
    int op, a, b, c;
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d%d%d", &a, &b, &c);
      double p = 1.0 * b / (1.0 * c);
      pair<double, double> D = make_pair((1 - p) / p, (1 - p) / p);
      update(a, a, D, 1);
    } else {
      scanf("%d%d", &a, &b);
      double p = query(a, b, 1).second;
      if (p < 1e20)
        printf("%.12f\n", 1.0 / (1.0 + p));
      else
        printf("0.000000000000\n");
    }
  }
  return 0;
}