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 = 100005;
double p[maxn];
struct st {
  double a;
  double b;
} seg[maxn * 4];
void pushup(int rt) {
  seg[rt].a = seg[rt * 2].a * seg[rt * 2 + 1].a;
  seg[rt].b = seg[rt * 2].b + seg[rt * 2].a * seg[rt * 2 + 1].b;
}
void build(int l, int r, int rt) {
  if (l == r) {
    seg[rt].a = (1 - p[l]) / p[l];
    seg[rt].b = (1 - p[l]) / p[l];
    return;
  }
  int mid = (l + r) / 2;
  build(l, mid, rt * 2);
  build(mid + 1, r, rt * 2 + 1);
  pushup(rt);
}
void update(int l, int r, int rt, int pos) {
  if (l == r && l == pos) {
    seg[rt].a = (1 - p[l]) / p[l];
    seg[rt].b = (1 - p[l]) / p[l];
    return;
  }
  int mid = (l + r) / 2;
  if (pos <= mid)
    update(l, mid, rt * 2, pos);
  else
    update(mid + 1, r, rt * 2 + 1, pos);
  pushup(rt);
}
double querya(int rt, int l, int r, int L, int R) {
  if (l >= L && r <= R) {
    return seg[rt].a;
  }
  int mid = (l + r) / 2;
  if (R <= mid) return querya(rt * 2, l, mid, L, R);
  if (L > mid) return querya(rt * 2 + 1, mid + 1, r, L, R);
  return querya(rt * 2, l, mid, L, R) * querya(rt * 2 + 1, mid + 1, r, L, R);
}
double queryb(int rt, int l, int r, int L, int R) {
  if (l >= L && r <= R) {
    return seg[rt].b;
  }
  int mid = (l + r) / 2;
  if (R <= mid) return queryb(rt * 2, l, mid, L, R);
  if (L > mid) return queryb(rt * 2 + 1, mid + 1, r, L, R);
  return queryb(rt * 2, l, mid, L, R) +
         querya(rt * 2, l, mid, L, R) * queryb(rt * 2 + 1, mid + 1, r, L, R);
}
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);
    p[i] = 1.0 * a / b;
  }
  build(1, n, 1);
  for (int i = 0; i < q; i++) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int x;
      int a, b;
      scanf("%d%d%d", &x, &a, &b);
      p[x] = 1.0 * a / b;
      update(1, n, 1, x);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      double qry = queryb(1, 1, n, l, r) + 1;
      printf("%.10f\n", 1.0 / qry);
    }
  }
  return 0;
}