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 N = 4e5 + 10;
const long long INF = 1e9 + 19;
struct Seg {
  double l, r;
  bool flag;
  Seg() {}
  Seg(double l, double r) : l(l), r(r), flag(0) {}
  Seg(bool flag) : flag(flag) {}
};
Seg merge(Seg w1, Seg w2) {
  if (w1.flag) return w2;
  if (w2.flag) return w1;
  double b = w2.l / (1 - (1 - w2.l) * w1.r);
  return Seg(w1.l * b, w2.r + (1 - w2.r) * w1.r * b);
}
int n, q;
double p[N];
Seg tree[N];
void build(int v, int ll, int rr) {
  if (ll + 1 == rr) {
    tree[v] = Seg(p[ll], p[ll]);
    return;
  }
  build(v * 2 + 1, ll, (ll + rr) / 2);
  build(v * 2 + 2, (ll + rr) / 2, rr);
  tree[v] = merge(tree[v * 2 + 1], tree[v * 2 + 2]);
}
void sett(int v, int ll, int rr, int pos, double val) {
  if (pos < ll || rr <= pos) return;
  if (ll + 1 == rr) {
    tree[v] = Seg(val, val);
    return;
  }
  sett(v * 2 + 1, ll, (ll + rr) / 2, pos, val);
  sett(v * 2 + 2, (ll + rr) / 2, rr, pos, val);
  tree[v] = merge(tree[v * 2 + 1], tree[v * 2 + 2]);
}
Seg get(int v, int ll, int rr, int l, int r) {
  if (l >= rr || ll >= r) return Seg(1);
  if (l <= ll && rr <= r) return tree[v];
  return merge(get(v * 2 + 1, ll, (ll + rr) / 2, l, r),
               get(v * 2 + 2, (ll + rr) / 2, rr, l, r));
}
void read() {
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    p[i] = x * 1.0 / y;
  }
}
void solve() {
  build(0, 0, n);
  for (int i = 0; i < q; i++) {
    int type;
    scanf("%d", &type);
    if (type == 2) {
      int l, r;
      scanf("%d%d", &l, &r);
      l--;
      Seg res = get(0, 0, n, l, r);
      printf("%.17f\n", res.l);
    } else {
      int pos, a, b;
      scanf("%d%d%d", &pos, &a, &b);
      sett(0, 0, n, pos - 1, a * 1.0 / b);
    }
  }
}
void stress() {}
int main() {
  if (1) {
    int k = 1;
    for (int tt = 0; tt < k; tt++) {
      read();
      solve();
    }
  } else {
    stress();
  }
  return 0;
}