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 + 10;
struct TreeNode {
  int L, R;
  pair<double, double> val;
} tree[maxn << 2];
int a[maxn], b[maxn], n, q;
void pushUp(int root) {
  tree[root].val.first =
      tree[root << 1].val.first * tree[root << 1 | 1].val.first;
  tree[root].val.second =
      tree[root << 1].val.second +
      tree[root << 1].val.first * tree[root << 1 | 1].val.second;
}
void build(int L, int R, int root) {
  tree[root].L = L;
  tree[root].R = R;
  if (L == R) {
    return;
  }
  int mid = (L + R) >> 1;
  build(L, mid, root << 1);
  build(mid + 1, R, root << 1 | 1);
}
void update(int pos, pair<double, double> val, int root) {
  int L = tree[root].L, R = tree[root].R;
  if (L == R) {
    tree[root].val = val;
    return;
  }
  int mid = (L + R) >> 1;
  if (pos <= mid) {
    update(pos, val, root << 1);
  } else {
    update(pos, val, root << 1 | 1);
  }
  pushUp(root);
}
pair<double, double> query(int ql, int qr, int root) {
  int L = tree[root].L, R = tree[root].R;
  if (L >= ql && R <= qr) return tree[root].val;
  int mid = (L + R) >> 1;
  pair<double, double> a = make_pair(1.0, 0.0), b = make_pair(1.0, 0.0), c;
  if (ql <= mid) {
    a = query(ql, qr, root << 1);
  }
  if (qr > mid) {
    b = query(ql, qr, root << 1 | 1);
  }
  c.first = a.first * b.first;
  c.second = a.second + a.first * b.second;
  return c;
}
int main() {
  cin >> n >> q;
  build(1, n, 1);
  for (int i = 1; i <= n; i++) {
    double x, y;
    cin >> x >> y;
    double p = x / y;
    update(i, make_pair((1 - p) / p, (1 - p) / p), 1);
  }
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int i, x, y;
      scanf("%d%d%d", &i, &x, &y);
      double p = double(x) / y;
      update(i, make_pair((1.0 - p) / p, (1.0 - p) / p), 1);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      pair<double, double> t = query(l, r, 1);
      double p = t.second;
      if (p < 1e20)
        printf("%.6lf\n", 1.0 / (1 + p));
      else
        cout << "0.0000000" << endl;
    }
  }
  return 0;
}