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 = 100000;
int N;
double T[Maxn + 5];
struct SegmentTree {
  struct Segment {
    double mul, sum;
  };
  Segment t[Maxn * 4 + 5];
  void pushup(int rt) {
    t[rt].mul = t[rt << 1].mul * t[rt << 1 | 1].mul;
    t[rt].sum = t[rt << 1].sum + t[rt << 1].mul * t[rt << 1 | 1].sum;
  }
  void build(int rt, int l, int r) {
    if (l == r) {
      t[rt].mul = t[rt].sum = T[l];
      return;
    }
    int mid = (l + r) >> 1;
    build(rt << 1, l, mid), build(rt << 1 | 1, mid + 1, r);
    pushup(rt);
  }
  void modify(int rt, int l, int r, int pos, double val) {
    if (l == r) {
      t[rt].mul = t[rt].sum = val;
      return;
    }
    int mid = (l + r) >> 1;
    if (pos <= mid)
      modify(rt << 1, l, mid, pos, val);
    else
      modify(rt << 1 | 1, mid + 1, r, pos, val);
    pushup(rt);
  }
  double __query(int rt, int l, int r, int ql, int qr) {
    if (ql <= l && r <= qr) return t[rt].mul;
    int mid = (l + r) >> 1;
    double ret = 1.0;
    if (ql <= mid) ret *= __query(rt << 1, l, mid, ql, qr);
    if (qr > mid) ret *= __query(rt << 1 | 1, mid + 1, r, ql, qr);
    return ret;
  }
  double query(int rt, int l, int r, int ql, int qr) {
    if (ql <= l && r <= qr) return t[rt].sum;
    int mid = (l + r) >> 1;
    double ret = 0;
    if (ql <= mid) ret += query(rt << 1, l, mid, ql, qr);
    if (qr > mid)
      ret += __query(rt << 1, l, mid, ql, mid) *
             query(rt << 1 | 1, mid + 1, r, ql, qr);
    return ret;
  }
};
SegmentTree tree;
int main() {
  int Q;
  scanf("%d %d", &N, &Q);
  for (int i = 1; i <= N; i++) {
    double a, b;
    scanf("%lf %lf", &a, &b);
    T[i] = (b - a) / a;
  }
  tree.build(1, 1, N);
  while (Q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int pos;
      double a, b;
      scanf("%d %lf %lf", &pos, &a, &b);
      tree.modify(1, 1, N, pos, (b - a) / a);
    }
    if (op == 2) {
      int l, r;
      scanf("%d %d", &l, &r);
      printf("%.10f\n", 1 / (tree.query(1, 1, N, l, r) + 1.0));
    }
  }
  return 0;
}