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;
int n, m;
double a[100005];
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  return x * f;
}
inline void print(int x) {
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
struct segment_tree {
  struct treenode {
    double x, y;
    treenode() {}
    treenode(double t) { x = y = t; }
  } tree[4 * 100005], ans;
  treenode merge(treenode ls, treenode rs) {
    treenode ans(0);
    double l1 = ls.x, l2 = rs.x, r1 = ls.y, r2 = rs.y;
    ans.x = l1 * l2 / (1 - (1 - l2) * r1),
    ans.y = l2 * (1 - r2) * r1 / (1 - (1 - l2) * r1) + r2;
    return ans;
  }
  void build(int p, int l, int r) {
    if (l == r) {
      tree[p].x = tree[p].y = a[l];
      return;
    }
    int mid = (l + r) >> 1;
    build(p << 1, l, mid), build(p << 1 | 1, mid + 1, r);
    tree[p] = merge(tree[p << 1], tree[p << 1 | 1]);
  }
  void change(int p, int l, int r, int pos, double val) {
    if (l == r) {
      tree[p].x = tree[p].y = val;
      return;
    }
    int mid = (l + r) >> 1;
    if (pos <= mid)
      change(p << 1, l, mid, pos, val);
    else
      change(p << 1 | 1, mid + 1, r, pos, val);
    tree[p] = merge(tree[p << 1], tree[p << 1 | 1]);
  }
  void query(int p, int l, int r, int x, int y) {
    if (x <= l && r <= y) {
      if (ans.x == -1)
        ans = tree[p];
      else
        ans = merge(ans, tree[p]);
      return;
    }
    int mid = (l + r) >> 1;
    if (x <= mid) query(p << 1, l, mid, x, y);
    if (y > mid) query(p << 1 | 1, mid + 1, r, x, y);
  }
} T;
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; i++) {
    int x = read(), y = read();
    a[i] = 1.0 * x / y;
  }
  T.build(1, 1, n);
  while (m--) {
    int type = read();
    if (type == 1) {
      int pos = read(), x = read(), y = read();
      T.change(1, 1, n, pos, 1.0 * x / y), a[pos] = 1.0 * x / y;
    } else {
      int l = read(), r = read();
      T.ans.x = T.ans.y = -1;
      T.query(1, 1, n, l, r);
      printf("%lf\n", T.ans.x);
    }
  }
  return 0;
}