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;
template <class T>
void read(T &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  x *= f;
}
int n, m;
double jl[100005], tree1[400005], tree2[400005];
void build(int l, int r, int rt) {
  if (l == r) {
    tree1[rt] = jl[l];
    tree2[rt] = jl[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, rt << 1);
  build(mid + 1, r, rt << 1 | 1);
  tree1[rt] = tree1[rt << 1] * tree1[rt << 1 | 1];
  tree2[rt] = tree2[rt << 1] + 1.0 * (tree2[rt << 1 | 1]) * tree1[rt << 1];
}
void update(int to, double x, int l, int r, int rt) {
  if (l == r) {
    tree1[rt] = x;
    tree2[rt] = x;
    return;
  }
  int mid = (l + r) >> 1;
  if (to <= mid)
    update(to, x, l, mid, rt << 1);
  else
    update(to, x, mid + 1, r, rt << 1 | 1);
  tree1[rt] = tree1[rt << 1] * tree1[rt << 1 | 1];
  tree2[rt] = tree2[rt << 1] + 1.0 * (tree2[rt << 1 | 1]) * tree1[rt << 1];
}
double query1(int L, int R, int l, int r, int rt) {
  if (L <= l && R >= r) return tree1[rt];
  int mid = (l + r) >> 1;
  double ans = 1.0;
  if (L <= mid) ans = 1.0 * ans * query1(L, R, l, mid, rt << 1);
  if (R > mid) ans = 1.0 * ans * query1(L, R, mid + 1, r, rt << 1 | 1);
  return ans;
}
double query(int L, int R, int l, int r, int rt) {
  if (L <= l && R >= r) return tree2[rt];
  int mid = (l + r) >> 1;
  double ans = 0.0;
  if (L <= mid) ans += 1.0 * query(L, R, l, mid, rt << 1);
  if (R > mid)
    ans += 1.0 * query(L, R, mid + 1, r, rt << 1 | 1) *
           query1(L, mid, l, mid, rt << 1);
  return ans;
}
int main() {
  int a, b, c, qwq;
  read(n);
  read(m);
  for (int i = 1; i <= n; ++i) {
    read(a);
    read(b);
    jl[i] = 1.0 * (b - a) / a;
  }
  build(1, n, 1);
  while (m--) {
    read(qwq);
    if (qwq == 1) {
      read(a);
      read(b);
      read(c);
      jl[a] = 1.0 * (c - b) / b;
      update(a, jl[a], 1, n, 1);
    }
    if (qwq == 2) {
      read(a);
      read(b);
      printf("%.10f\n", 1.0 / (query(a, b, 1, n, 1) + 1.0));
    }
  }
}