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 MAX = 1e5 + 10;
const int MOD = 1e9 + 7;
double segta[4 * MAX];
double segtb[4 * MAX];
double prob[MAX];
void build(int pos, int l, int r) {
  if (l == r) {
    segta[pos] = prob[l];
    segtb[pos] = prob[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(pos + pos, l, mid);
  build(pos + pos + 1, mid + 1, r);
  segta[pos] = (segta[pos + pos] * segta[pos + pos + 1]);
  segtb[pos] = segtb[pos + pos] + (segta[pos + pos] * segtb[pos + pos + 1]);
}
void upd(int pos, int l, int r, int ind, double val) {
  if (l > ind || r < ind) {
    return;
  }
  if (l == r && l == ind) {
    segta[pos] = (1.0 - val) / val;
    segtb[pos] = (1.0 - val) / val;
    return;
  }
  int mid = (l + r) >> 1;
  upd(pos + pos, l, mid, ind, val);
  upd(pos + pos + 1, mid + 1, r, ind, val);
  segta[pos] = (segta[pos + pos] * segta[pos + pos + 1]);
  segtb[pos] = segtb[pos + pos] + (segta[pos + pos] * segtb[pos + pos + 1]);
}
pair<double, double> query(int pos, int l, int r, int ql, int qr) {
  if (ql > r || qr < l) {
    return make_pair(0.0, 1.0);
  }
  if (ql <= l && qr >= r) {
    return make_pair(segtb[pos], segta[pos]);
  }
  int mid = (l + r) >> 1;
  pair<double, double> tmp1 = query(pos + pos, l, mid, ql, qr);
  pair<double, double> tmp2 = query(pos + pos + 1, mid + 1, r, ql, qr);
  double help1 = (tmp1.second * tmp2.second);
  double help2 = tmp1.first + (tmp2.first * tmp1.second);
  return make_pair(help2, help1);
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    prob[i] = (double)a / b;
    prob[i] = (1.0 - prob[i]) / prob[i];
  }
  build(1, 1, n);
  for (int i = 0; i < q; i++) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int a, b, c;
      scanf("%d %d %d", &a, &b, &c);
      upd(1, 1, n, a, (double)b / c);
    } else {
      int a, b;
      scanf("%d %d", &a, &b);
      double val = query(1, 1, n, a, b).first;
      if (val < 1e30) {
        printf("%lf\n", 1 / (1.0 + val));
      } else {
        printf("0.0000\n");
      }
    }
  }
}