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;
vector<double> parr;
struct Node {
  double p, q;
};
struct BIT {
  vector<Node> tree;
  BIT(int N) { tree = vector<Node>(4 * (N + 1)); }
  void build(int node, int l, int r) {
    if (l == r) {
      tree[node].p = parr[l];
      tree[node].q = 1.0 - parr[l];
      return;
    }
    int m = (l + r) >> 1;
    build(node * 2, l, m), build(node * 2 + 1, m + 1, r);
    tree[node] = merge(tree[node * 2], tree[node * 2 + 1]);
  }
  void update(int node, int l, int r, int idx, double v) {
    if (idx < l || r < idx) return;
    if (idx <= l && r <= idx) {
      tree[node].p = v;
      tree[node].q = 1.0 - v;
      return;
    }
    int m = (l + r) >> 1;
    update(node * 2, l, m, idx, v), update(node * 2 + 1, m + 1, r, idx, v);
    tree[node] = merge(tree[node * 2], tree[node * 2 + 1]);
  }
  Node merge(Node& l, Node& r) {
    double pl = l.p, pr = r.p;
    double ql = l.q, qr = r.q;
    Node ret;
    ret.p = pl * pr / (1.0 - (1.0 - pr) * (1.0 - ql));
    ret.q = qr * ql / (1.0 - (1.0 - ql) * (1.0 - pr));
    return ret;
  }
  Node query(int node, int l, int r, int a, int b) {
    if (r < a || b < l) return {1.0, 1.0};
    if (a <= l && r <= b) {
      return tree[node];
    }
    int m = (l + r) >> 1;
    Node L = query(node * 2, l, m, a, b);
    Node R = query(node * 2 + 1, m + 1, r, a, b);
    return merge(L, R);
  }
};
int main() {
  int N, Q;
  scanf("%d%d", &N, &Q);
  parr.resize(N + 1);
  for (int i = 1, a, b; i <= N; i++) {
    scanf("%d%d", &a, &b);
    parr[i] = (double)a / (double)b;
  }
  BIT bit(N);
  bit.build(1, 1, N);
  for (int i = 1, type; i <= Q; i++) {
    scanf("%d", &type);
    if (type == 2) {
      int l, r;
      scanf("%d%d", &l, &r);
      printf("%0.6f\n", bit.query(1, 1, N, l, r).p);
    } else {
      int idx, a, b;
      scanf("%d%d%d", &idx, &a, &b);
      bit.update(1, 1, N, idx, (double)a / (double)b);
    }
  }
  return 0;
}