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, Q, A[(400010)], B[(400010)], lef[(400010)];
long double g[(400010)], f[(400010)];
struct node {
  long double f, g;
};
inline int gi() {
  int f = 1, ret = 0;
  char ch;
  do ch = getchar();
  while (!(ch >= '0' && ch <= '9') && ch != '-');
  if (ch == '-') f = -1, ch = getchar();
  do ret = ret * 10 + ch - '0', ch = getchar();
  while (ch >= '0' && ch <= '9');
  return f * ret;
}
inline void build(int now, int l, int r) {
  if (l == r) {
    lef[l] = now;
    g[now] = f[now] = (long double)A[l] / (long double)B[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(now << 1, l, mid);
  build(now << 1 | 1, mid + 1, r);
  f[now] = (f[now << 1] * f[now << 1 | 1]) /
           (1 - g[now << 1] * (1 - f[now << 1 | 1]));
  g[now] = g[now << 1 | 1] + (1 - g[now << 1 | 1]) * g[now << 1] *
                                 f[now << 1 | 1] /
                                 (1 + (f[now << 1 | 1] - 1) * g[now << 1]);
}
inline node query(int now, int l, int r, int ql, int qr) {
  if (l == ql && r == qr) return (node){f[now], g[now]};
  int mid = (l + r) >> 1;
  if (qr <= mid)
    return query(now << 1, l, mid, ql, qr);
  else if (ql > mid)
    return query(now << 1 | 1, mid + 1, r, ql, qr);
  else {
    node a, b, ret;
    a = query(now << 1, l, mid, ql, mid);
    b = query(now << 1 | 1, mid + 1, r, mid + 1, qr);
    ret.f = (a.f * b.f) / (1 - a.g * (1 - b.f));
    ret.g = b.g + (1 - b.g) * a.g * b.f / (1 + (b.f - 1) * a.g);
    return ret;
  }
}
int main() {
  scanf("%d %d", &N, &Q);
  for (int i = 1; i <= N; ++i) A[i] = gi(), B[i] = gi();
  build(1, 1, N);
  while (Q--) {
    int opt = gi();
    if (opt == 1) {
      int now = lef[gi()], a = gi(), b = gi();
      f[now] = g[now] = (long double)a / (long double)b;
      for (now >>= 1; now; now >>= 1) {
        f[now] = (f[now << 1] * f[now << 1 | 1]) /
                 (1 - g[now << 1] * (1 - f[now << 1 | 1]));
        g[now] = g[now << 1 | 1] +
                 (1 - g[now << 1 | 1]) * g[now << 1] * f[now << 1 | 1] /
                     (1 + (f[now << 1 | 1] - 1) * g[now << 1]);
      }
    } else {
      int l = gi(), r = gi();
      printf("%.10lf\n", (double)query(1, 1, N, l, r).f);
    }
  }
  return 0;
}